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yawa3891 [41]
3 years ago
6

I need help with this please

Mathematics
1 answer:
givi [52]3 years ago
8 0
C is 20 degrees. Pls mark brainliest
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If f(x) = |-3x - 6| - 10, find f(5)
brilliants [131]
| -3(5) - 6 |
-15 - 6
21 - 10
=11
I made the 21 positive because when it’s in the absolute value bracket it’s ALWAYS positive. Any thing inside the bracket is positive. So I think the answer 11.
6 0
3 years ago
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Heights of adult women in the united states are normally distributed with a population mean of μ = 63.5 inches and a population
maw [93]
Given:
<span>population mean of μ = 63.5 inches
population standard deviation of σ = 2.5 inches
</span><span>adult woman who has a height of 50.4 inches
</span>
<span>z = (x – μ) / σ

z = (50.4 - 63.5) / 2.5 = -13.1 / 2.5 = -5.24 

The z-value of an adult woman who has a height of 50.4 inches is -5.24 or -5.24 standard deviation below the mean. </span>
3 0
3 years ago
What is the value of a?<br> 45°<br> (2x-5)
aniked [119]

Answer:

x=70

Step-by-step explanation:

To solve this problem, you set up the equation 45+(2x-5)=180

Next, subtract 45 from 180 to get the equation 2x-5=135

After that, add 5 to 135 to get 2x=140

Divide 140 by 2 to get X=70

7 0
2 years ago
Solve for x:a(a²+b²)x²+b²x-a​
m_a_m_a [10]

Answer:

x = a/(a² + b²) or x = -1/a  

Step-by-step explanation:

a(a²+ b²)x² + b²x - a =0

Use the quadratic equation formula:

x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a} =\dfrac{-b\pm\sqrt{D}}{2a}

1. Evaluate the discriminant D

D = b² - 4ac = b⁴ - 4a(a² + b²)(-a) = b⁴ + 4a⁴ + 4a²b²  = (b² + 2a²)²

2. Solve for x

\begin{array}{rcl}x & = & \dfrac{-b\pm\sqrt{D}}{2a}\\\\ & = & \dfrac{-b^{2}\pm\sqrt{(b^{2}+2a^{2})^{2}}}{2a(a^{2} + b^{2})}\\\\ & = & \dfrac{-b^{2}\pm(b^{2}+ 2a^{2})}{2a(a^{2} + b^{2})}\\\\x = \dfrac{-b^{2}+(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}&\qquad& x =\dfrac{-b^{2}-(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}\\\\x =\dfrac{-b^{2}+(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}&\qquad& x =\dfrac{-b^{2}-(b^{2} +2a^{2})}{2a(a^{2} + b^{2})}\\\\\end{array}

\begin{array}{rcl}x = \large \boxed{\mathbf{\dfrac{a}{a^{2} + b^{2}}}}&\qquad& x =\dfrac{-b^{2}-(b^{2} +2a^{2})}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\dfrac{-2b^{2}- 2a^{2}}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\dfrac{-2(a^{2}+ b^{2})}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\large \boxed{\mathbf{-\dfrac{1}{a}}}\\\\\end{array}

5 0
3 years ago
Find p and q for which the linear eqn has infinite solutions
jekas [21]

Answer:

p= 2.5

q= 7

Step-by-step explanation:

The lines should overlap to have infinite solutions, slopes should be same and y-intercepts should be same.

Equations in slope- intercept form:

6x-(2p-3)y-2q-3=0 ⇒ (2p-3)y= 6x -2q-3 ⇒ y= 6/(2p-3)x -(2q+3)/(2p-3)

12x-( 2p-1)y-5q+1=0 ⇒ (2p-1)y= 12x - 5q+1 ⇒  y=12/(2p-1)x - (5q-1)/(2p-1)

Slopes equal:

6/(2p-3)= 12/(2p-1)

6(2p-1)= 12(2p-3)

12p- 6= 24p - 36

12p= 30

p= 30/12

p= 2.5

y-intercepts equal:

(2q+3)/(2p-3)= (5q-1)/(2p-1)

(2q+3)/(2*2.5-3)= (5q-1)/(2*2.5-1)

(2q+3)/2= (5q-1)/4

4(2q+3)= 2(5q-1)

8q+12= 10q- 2

2q= 14

q= 7

7 0
3 years ago
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