Answer:
Sqrt (11/8)
Step-by-step explanation:
Sec^2 - Tan^2 = 1
- Tan^2 = 1 - Sec^2
Tan^2 = -1 + Sec^2
Tan^2 = 3/8
3/8 = -1 + Sec^2 Theta
1 + 3/8 = Sec ^2 Theta
11/8 = Sec^2 Theta
Sqrt ( 11/8) = Sec of theta
-- Gage Millar, Algebra 1/2 Tutor (Pending Pre-Calc Tutor)
Given that 24w = 14ten .Find the unknown base
Answer:
w = 5
Step-by-step explanation:
24w = (14)10
2w + 4 = 10 + 4
2w = 10
w = 5
solve 2(r+6) =(6+1/3r)
Answer:
r = - 18/5
I hope this helps! ^^
( This problem has 13 steps, so it would take a lot of time to type them in ^^)
F. In what order should items weighing
51 pounds, 40 pounds, 48 pounds,
and 44 pounds be stacked if you want
them in order from heaviest to lightest?
Answer:
51 pounds, 48 pounds, 44 pounds and 40 pounds in that chronological order.
what are the soltuions to the quadratic equation below? 12x squared + 4x -5=0
Answer: 0.5 or - 0.834
Step-by-step explanation: Here is the explanation!
Solve for X.
6
X = [?]
Enter the number, in decimal form,
that belongs in the green box.
Answer:
By proportional method
[tex] \frac{4}{6} = \frac{5}{x} [/tex]
4x= 30
x= 7.5
what is the area of the figure?
Answer:
area = 62ft
Step-by-step explanation
so you section the polygon into a big rectangle and 2 small rectangles
the area of the small rectangle will be 2ft x 3ft which will be 6 ft, and as there are 2 of them the area of it will be 12ft
the area of the big rectangle will be 10ft x 5ft, as 8ft - 3ft is 5ft so 5x10=50ft
now add the areas you got, 50+12 which is 62ft
hope this helps:)
I need a little help here
WILL MARK BRAINLIEST 50 POINTS Find the area of the regular pentagon if the apothem is 7 ft and a side is 10 ft. Round to the nearest whole number.
175 ft2
350 ft2
35 ft2
70 ft2
Answer:
175 ft^2
Step-by-step explanation:
split the pentagon into 5 triangles with base length 10ft and height 7ft. each triangle then has an area of 10 * 7 * 1/2 = 35 ft^2
then the pentagon has an area 35*5 = 175 ft^2
-3+8x-5=-8 solve for x
The value of this equation is x = 0
The representation of an equation of the first degree - is given by:
[tex] \boxed{ \large \sf ax + b = 0}[/tex]
This representation is also defined in a first degree function.
— To solve this expression, let's: add or subtract the real terms. Then we will do the division.
⚘ Resolution
[tex] \large \sf{-3+8x-5=-8}[/tex]
[tex] \large \sf{-8+8x=-8}[/tex]
[tex] \large \sf{8x=0}[/tex]
[tex] \large \sf{x=0 \div 8}[/tex]
[tex] \green{ \boxed{ \boxed{ \blue{ \large \sf{x=0}}}}} \\ [/tex]
Therefore, the value of this equation will be x = 0
can someone help me with this worksheet please!!!!
(1) The missing term in the sequence, a₁₂ = 0.8.
(2) The missing term in the sequence, a₈ = 102.5.
(3) The missing term in the sequence, a₈ = 111.
(4) The missing term in the sequence, a₁₂ = -19.
(5) The missing term in the sequence, a₁₂ = 94.
(6) The missing term in the sequence, a₆ = 40.
(7) The missing term in the sequence, a₃₆ = -52.
(8) The missing term in the sequence, a₂₁ = -58.
Missing term of the sequenceThe missing term in the sequence is determined as follows;
Tₙ = a + (n - 1)d
1.0 a₄ = 18.4 and a₅ = 16.2, a₁₂ = ?T₄ = a + 3d
18.4 = a + 3d ---(1)
T₅ = a + 4d
16.2 = a + 4d ---(2)
subtract (1) from (2)
-2.2 = d
18.4 = a + 3(-2.2)
a = 25
a₁₂ = a + 11d
a₁₂ = 25 + 11(-2.2)
a₁₂ = 0.8
2.0 a₂ = 57.5 and a₅ = 80, a₈ = ?a₂ = a + d
57.5 = a + d -- (1)
a₅ = a + 4d
80 = a + 4d --- (2)
solve (1) and (2)
d = 7.5
a = 50
a₈ = a + 7d
a₈ = 50 + 7(7.5)
a₈ = 102.5
3.0 a₁₀ = 141 and a₁₃ = 186, a₈ = ?a₁₀ = a + 9d
141 = a + 9d --- (1)
a₁₃ = a + 12d
186 = a + 12d --- (2)
Subtract (1) from (2)
d = 15
a = 6
a₈ = a + 7d
a₈ = 6 + 7(15)
a₈ = 111
4.0 a₂₂ = -49 and a₂₅ = -58, a₁₂ = ?a₂₂ = a + 21d
-49 = a + 21d ---- (1)
a₂₅ = a + 24d
-58 = a + 24d --- (2)
subtract (1) from (2)
d = -3
a = 14
a₁₂ = a + 11d
a₁₂ = 14 + 11(-3)
a₁₂ = -19
5.0 a₄ = -2 and a₈ = 46, a₁₂ = ?a₄ = a + 3d
-2 = a + 3d --- (1)
a₈ = a + 7d
46 = a + 7d ---- (2)
Subtract (1) from (2)
d = 12
a = -38
a₁₂ = a + 11d
a₁₂ = -38 + 11(12)
a₁₂ = 94
6.0 a₉ = 64 and a₁₂ = 88, a₆ = ?a₉ = a + 8d
64 = a + 8d --- (1)
a₁₂ = a + 11d
88 = a + 11d --- (2)
Subtract (1) from (2)
d = 8
a = 0
a₆ = a + 5d
a₆ = 0 + 5(8)
a₆ = 40
7.0 a₂₀ = -4 and a₂₃ = -13, a₃₆ = ?a₂₀ = a + 19d
-4 = a + 19d ---- (1)
a₂₃ = a + 22d
-13 = a + 22d --- (2)
Subtract (1) from (2)
d = -3
a = 53
a₃₆ = a + 35d
a₃₆ = 53 + 35(-3)
a₃₆ = -52
8.0 a₂₈ = 5 and a₃₃ = 50, a₂₁ = ?a₂₈ = a + 27d
5 = a + 27d ---- (1)
a₃₃ = a + 32d
50 = a + 32d --- (2)
Subtract (1) from (2)
d = 9
a = -238
a₂₁ = a + 20d
a₂₁ = -238 + 20(9)
a₂₁ = -58
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find the area of the shaded polygons
Answer:
4 square units
Step-by-step explanation:
The vertices of the figure are on grid points, so it is appropriate to use Pick's theorem to find the area.
__
formulaPick's theorem tells you the area is ...
A = i +b/2 -1
where i is the number of grid points interior to the figure (0), and b is the number of grid points on the boundary (10).
applicationUsing the counted values in the formula, we find the area to be ...
A = 0 +10/2 -1 = 4
The area of the polygon is 4 square units.
_____
Additional comment
There are several other ways to find the area. Here are a couple:
decompose the figure
A horizontal line 1 unit up from the bottom will divide the figure into a trapezoid and a triangle. The trapezoid has bases 4 and 1, and height 1, so its area is ...
A = 1/2(b1 +b2)h = 1/2(4 +1)(1) = 5/2
The triangle has base 1 and height 3, so its area is ...
A = 1/2bh = 1/2(1)(3) = 3/2
Then the total area is 5/2 +3/2 = 8/2 = 4 square units.
subtract empty space
The figure occupies a 4×4 square with triangles removed from the left side and the top. Each of those triangles has a base of 4 and a height of 3. The remaining (shaded) area is ...
A = s² -1/2bh -1/2bh
A = 4² -1/2(4)(3) -1/2(4)(3) = 16 -12 = 4 square units
Simplify the expression.
Show your work.
4 √81m^8n^16
Answer:
hope you can understand
Answer:
3m²n⁴
Step-by-step explanation:
Given :
⇒ [tex]\sqrt[4]{81m^{8}n^{16} }[/tex]
=============================================================
Solving :
⇒ [tex]\sqrt[4]{81}[/tex] × [tex]\sqrt[4]{m^{8} }[/tex] × [tex]\sqrt[4]{n^{16} }[/tex]
⇒ [tex]\sqrt[4]{3^{4} }[/tex] × [tex]\sqrt[4]{(m^{2})^{4} }[/tex] × [tex]\sqrt[4]{(n^{4})^{4} }[/tex]
⇒ 3 × m² × n⁴
⇒ 3m²n⁴
x − y = 12
x + 2y = 21
Answer: x=15, y=3
Step-by-step explanation:
Subtracting the two equations, we get -3y=-9, meaning y=3.
Substituting this into the first equation, we get that x-3=12, and thus, x=15.
please help 35 points!!
Answer:
67m²
Step-by-step explanation:
12m + 6m + 4m + 36 m 8m = 67m²
Select ALL the correct answers. Consider the following graph of function f. Which transformations will change function f into function g given below. a vertical shift down 3 units a vertical shift down 5 units a vertical shift up 5 units a horizontal shift left 7 units a horizontal shift right 7 units a horizontal shift left 4 units
In triangle ABC, AP is an angle bisector of angle BAC. What is the length of PC? Round you answer to the nearest whole number.
A) 6 B) 7 C) 8 D) 9
Answer:
D) 9
Step-by-step explanation:
x = ½ × 13 = 6.5
y = ½ × 5 = 2.5
6.5 + 2.5 = 9
So the length of PC is 9
HOPE THIS HELPS AND HAVE A NICE DAY <3
find the exact value of sin 15 degrees
Answer:
Hi,
Step-by-step explanation:
sin(a-b)=sin(a) cos(b)+ cos(a) sin(b)
a=45° and b=30°
[tex]sin(45^o-30^o)=sin(45^o)*cos(30^o)+cos(45^o)*sin(30^o)\\\\=\dfrac{\sqrt{2}}{2}* \dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{2}}{2}*\dfrac{1}{2}\\\\\\\boxed{sin(15^o)=\dfrac{\sqrt{2}}{4}*(1+\sqrt{3} )}\\[/tex]
Find sin A for the triangle below. Give the exact value as an expression and an approximation to the nearest ten-thousandth. Note: The triangle is not drawn to scale.
the exact value of sin A to the nearest ten- thousandth is 0. 6625
Using the pythagorean theorem
a² + b² = c²
The opposite side is unknown, so use the pythagorean theorem to find it
c = hypotenuse = 4
a= opposite site = ?
b= adjacent side = 3
Substitute into the formula
4² = a² + 3²
16 = a² + 9
a² = 16 -9 = 7
Find the square root
a =√7 = 2. 65
To find Sin A, use
Sin A = opposite side ÷ hypotenuse
Sin A = 2. 65 ÷ 4 = 0. 6625
Thus, the value of sin A is 0. 6625
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What is the domain of the function shown in this graph?
Answer:
[tex] - \infty < x < \infty [/tex]
Step-by-step explanation:
On the right hand side of the y axis, the x values will continue to increase. They will go to positive Infinity. On the left hand side of the y axis, the x values will continue to decrease, this means that they will become more negative. They will go to negative Infinity.Write two numbers that multiply to the value on top and add to the value on bottom.
-84
17
Two numbers multiply the value on top (-84) and add to the value on the bottom (17) is 21 and -4.
What is multiplication?Multiplication is one of the four basic arithmetic operations, alongside addition, subtraction, and division. Multiplication essentially means the repeated addition.
The two given numbers are -84 and 17.
We need to write two numbers that multiply to the value on top (-84) and add to the value on the bottom (17).
If we multiply 21 and -4, we get the product as -84.
That is, 21×(-4)=-84
If we add 21 and -4, we get the sum as -17.
That is, 21+(-4)=21-4=17
Therefore, two numbers multiply the value on top (-84) and add to the value on the bottom (17) is 21 and -4.
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Please help quickly!!
The values of a, b and c from the given exponential function are 7, 9 and 4 respectively
Laws of indicesAccording to the exponential law of indices
[tex]\sqrt[c]{a^b}[/tex]
This can be written as;
[tex]\sqrt[c]{a^b}=a^{\frac{b}{c} }[/tex]
Given the exponential expression
[tex]7^\frac{9}{4}[/tex]
Compare with the original expression
a = 7, b = 9 and c = 4
Hence the values of a, b and c from the given exponential function are 7, 9 and 4 respectively
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A distribution has the five-number summary shown below. What is the
interquartile range (IQ) of this distribution?
Answer:
tiookvgvc. jbjvth kivtcth jjvf h. bkbgv
Answer:
The IQR of the given distribution is
Step-by-step explanation:
The given distribution has the five-number
28, 34, 43, 59, 62
Divide these numbers in two equal parts.
(28, 34), 43,( 59, 62)
Now divide each parenthesis in two equal parts.
(28), (34), 43,( 59), (62)
It means first quartile is the average of 28 and 34. Third quartile is the average of 59 and 62.
The interquartile range (IQR) of this distribution is
Therefore the IQR of the given distribution is 29.5.
find the solution set. 4x^2+x=3
Answer:
[tex]x=\frac{3}{4},\:x=-1[/tex]
Keys:
For this problem, you need the quadratic formula(listed below).
[tex]x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex][tex]1^a=1[/tex][tex]\sqrt[n]{a}^n=a[/tex]When you see ± in a quadratic equation, you must know there is going to be at least 2 solutions.
Step-by-step explanation:
solving for x₁ and x₂
[tex]4x^2+x=3\\4x^2+x-3=3-3\\4x^2+x-3=0\\x_{1,\:2}=\frac{-1\pm \sqrt{1^2-4\cdot 4\left(-3\right)}}{2\cdot 4}\\[/tex]
[tex]1^2=1\\=\sqrt{1-4\cdot \:4\left(-3\right)}\\=\sqrt{1+4\cdot \:4\cdot \:3}\\=\sqrt{1+48}\\=\sqrt{49}\\=\sqrt{7^2}\\\sqrt{7^2}=7\\=7[/tex]
[tex]x_{1,\:2}=\frac{-1\pm \:7}{2\cdot \:4}\\x_1=\frac{-1+7}{2\cdot \:4},\:x_2=\frac{-1-7}{2\cdot \:4}\\[/tex]
solve for x₁
[tex]\frac{-1+7}{2\cdot \:4}[/tex]
[tex]=\frac{6}{2\cdot \:4}[/tex]
[tex]=\frac{6}{8}[/tex]
[tex]= \frac{6\div2}{8\div2}[/tex]
[tex]=\frac{3}{4}[/tex]
solve for x₂
[tex]\frac{-1-7}{2\cdot \:4}[/tex]
[tex]=\frac{-8}{2\cdot \:4}[/tex]
[tex]=\frac{-8}{8}[/tex]
[tex]=-\frac{8}{8}[/tex]
[tex]=-1[/tex]
Hope this helps!
( PLEASE HELP WITH THIS QUESTION)
You are studying a single-celled organism under a microscope. Is it possible for this organism to be classified as fungi?
Answer: Yes it is possible
Example: Yeast is a single-celled fungus.
There are probably other types of fungus that are single-celled. However, some other fungi are multi-celled. You will likely need more information about the organism under the microscrope before you can classify it properly.
A scale model of a building is 3 inches tall. If the building is 90 feet tall, find the scale of the model.
The scale model of a building 3 inches tall if the building is 90 feet tall is 2:5
Scaling of measurementScaling is the way of enlarging or reducing the image of an object.
Given scale model of a building is 3 inches, if the building is 90feet tall, to write the scale;
Convert both measures to same units
Since 1 feet. = 12 in
90 feet = 90/12 in = 7.5 in
Take the ratio
3 : 7.5 = 30: 75
Reduce to lowest term
30: 75 = 2 : 5
Hence the scale model of a building 3 inches tall if the building is 90 feet tall is 2:5
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A researcher asks the first 10 people he meets at a shopping mall about their opinions on a certain brand of shampoo. This is an example of a ________________.
A. self-selecting sample
B. convenience sample
C. random sample
D. systematic sample
Answer:
i believe the answer is A
Step-by-step explanation:
I need help please , I don’t really know much about pre calculus
Answer:
The angle Sita here is called Amplitude
You purchase a new car today.the value of that car depreciates based on the function f(t)=12,000(0.96)^t, where t is measured in years after purchase. How much is the car worth after 3/2 years, rounded to the nearest dollar
Considering the definition of exponential function, the value of the car is 11,287.25 dollars.
What is exponential functionAn exponential function is one in which the independent variable x appears in the exponent and has a constant a as its base. Its expression is:
f(x)= aˣ
Being a a positive real, a > 0, and different from 1, a ≠ 1.
When 0 < a < 1, then the exponential function is a decreasing function and when a > 1, it is an increasing function.
What is the price of the carIn this case, the value of that car depreciates based on the function f(t)=12,000×[tex]0.96^{t}[/tex] where t is measured in years after purchase.
After 3/2 years, the value of the car is calculated as:
f(3/2)=12,000×[tex]0.96^{3/2}[/tex]
Solving:
f(3/2)= 11,287.25
Finally, the value of the car is 11,287.25 dollars.
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Draw and set up the integrals for the area enclosed by the y–axis, the curve y = (x + 1)1/2 and y = 2. Compute one of them.
Region II only please
If the definitions of type I and type II regions is the same as in the link provided, then as a type I region the integration domain is the set
[tex]R_{\rm I} = \left\{(x,y) \mid 0 \le x \le 3 \text{ and } \sqrt{x+1} \le y \le 2\right\}[/tex]
and as a type II region,
[tex]R_{\rm II} = \left\{(x,y) \mid 0 \le x \le y^2-1 \text{ and } 1 \le y \le 2\right\}[/tex]
where we solve y = √(x + 1) for x to get x as a function of y.
A. The area of the type I region is
[tex]\displaystyle \iint_{R_{\rm I}} dA = \int_0^3 \int_{\sqrt{x+1}}^2 dy \, dx = \int_0^3 (2 - \sqrt{x+1}) \, dx = \boxed{\frac43}[/tex]
B. The area of the type II region is of course also
[tex]\displaystyle \iint_{R_{\rm II}} dA = \int_1^2 \int_0^{y^2-1} dx \, dy = \int_1^2 (y^2-1) \, dy = \boxed{\frac43}[/tex]
I've attached a plot of the type II region to give an idea of how it was determined. The black arrows indicate the domain of x as it varies from the line x = 0 (y-axis) to the curve y = √(x + 1).
im lost can someone help?