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2 years ago

6

The equation y = negative StartFraction 1 Over 20 EndFraction x + 10 represents the gallons of gasoline that remain in Michelle’

s car after she drives x miles. Which graph represents this equation?

2 answers:

klio [65]2 years ago

4 0

Answer: graph A

Step-by-step explanation: I took the test on ed genuity

andrey2020 [161]2 years ago

3 0

The graph must look like this (see attached):

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The product of sin 30 degrees and sin 60 degrees is the same as the product of

zysi [14]

Answer:

$sin(30\°)*sin(60\°)=cos(60\°)*cos(30\°)$

Step-by-step explanation:

we know that

$sin(\alpha)=cos(\beta)$

when

$\alpha+\beta=90\°$ -------> complementary angles

so

$sin(30\°)=cos(60\°)$

$sin(60\°)=cos(30\°)$

Because

$30\°+60\°=90\°$ -------> complementary angles

In this problem the product of

$sin(30\°)*sin(60\°)$

is equal to

$cos(60\°)*cos(30\°)$

therefore

$sin(30\°)*sin(60\°)=cos(60\°)*cos(30\°)$

3 0

2 years ago

Read 2 more answers

Which transformations have been applied to the graph of f(x) = x2 to produce the graph of g(x) = –5x2 + 100x – 450? Select three

RSB [31]

Answer:

The graph of f(x) is shifted up 50 units

The graph of f(x) is shifted right 10 units

The graph of f(x) is reflected over the x-axis

Step-by-step explanation:

we have

$f(x)=x^{2}$

This is a vertical parabola open upward

The vertex is a minimum

The vertex is the origin (0,0)

$g(x)=-5x^{2}+100x-450$

This is a vertical parabola open downward

The vertex is a maximum

The first thing to note is that fx) is a parabola that opens up and g(x) opens down, so a reflection across the x-axis must have been applied.

Find the vertex of g(x)

Convert to vertex form

$g(x)=-5x^{2}+100x-450$

Complete the square

$g(x)=-5(x^{2}-20x)-450$

$g(x)=-5(x^{2}-20x+100)-450+500$

$g(x)=-5(x^{2}-20x+100)+50$

$g(x)=-5(x-10)^{2}+50$

The vertex is the point (10,50)

so

To translate the vertex of (0,0) to (10,50)

The rule of the translation is

(x,y) ------> (x+10,y+50)

That means ----> The translation is 10 units at right and 50 units up

therefore

The transformations are

The graph of f(x) is shifted up 50 units

The graph of f(x) is shifted right 10 units

The graph of f(x) is reflected over the x-axis

5 0

2 years ago

Read 2 more answers

If ΔOCD ≅ ΔJAL, then which of the following statements are true? Select True or False for each statement.

andrey2020 [161]

Answer:

true

true

false

true

Step-by-step explanation:

4 0

2 years ago

R+5/mn=p solve for m

AnnZ [28]

Answer:

$\bold{m\ =\ \dfrac5{(p-r)n}}$

Step-by-step explanation:

$\bold{r+\dfrac5{mn}\ =\ p}\\\\ {}\quad-r\qquad-r\\\\{}\ \ \bold{\dfrac5{mn}\ =\ p-r}\\\\{}\ ^{_\times}(mn)\quad ^{_\times}(mn)\\\\{}\quad\bold{5\ =\ (p-r)^{_\times}(mn)}\\\\\div(p-r)\quad\div(p-r)\\\\{}\ \ \bold{\dfrac5{p-r}\ =\ mn}\\\\{}\quad \ \div n\quad\ \ \div n\\\\\bold{\dfrac5{(p-r)n}\ =\ m}$

If you mean (r+5)/mn then:

$\bold{\dfrac{r+5}{mn}\ =\ p}\\\\{}\ ^{_\times}(mn)\quad ^{_\times}(mn)\\\\{}\ \bold{r+5\ =\ pmn}\\\\\div(pn)\quad\div(pn)\\\\{}\ \ \bold{\dfrac{r+5}{pn}\ =\ m}$

4 0

2 years ago

Prove that (sec 12A-1)/(sec 6A-1)=tan 12A/tan 3A

adelina 88 [10]

Let $x=3A$ . Recall the following identities,

$\cos^2\theta=\dfrac{1+\cos2\theta}2$

$\sin^2\theta=\dfrac{1-\cos2\theta}2$

$\sin2\theta=2\sin\theta\cos\theta$

Now,

$\dfrac{\sec12A-1}{\sec6A-1}=\dfrac{\sec4x-1}{\sec2x-1}$

$=\dfrac{\cos2x(1-\cos4x)}{\cos4x(1-\cos2x)}$

$=\dfrac{2\cos2x\sin^22x}{\cos4x(1-\cos2x)}$

$=\dfrac{2\cos2x\sin^22x(1+\cos2x)}{\cos4x(1-\cos^22x)}=\dfrac{2\cos2x\sin^22x(1+\cos2x)}{\cos4x\sin^22x}=\dfrac{2\cos2x(1+\cos2x)}{\cos4x}$

$=\dfrac{4\cos2x\cos^2x}{\cos4x}$

$=\dfrac{4\cos2x\cos^2x\sin4x}{\cos4x\sin4x}=\dfrac{4\cos2x\cos^2x\tan4x}{\sin4x}$

$=\dfrac{4\cos2x\cos^2x\tan4x}{2\sin2x\cos2x}=\dfrac{2\cos^2x\tan4x}{\sin2x}$

$=\dfrac{2\cos^2x\tan4x}{2\sin x\cos x}=\dfrac{\cos x\tan4x}{\sin x}$

$=\dfrac{\tan4x}{\frac{\sin x}{\cos x}}=\dfrac{\tan4x}{\tan x}=\dfrac{\tan12A}{\tan3A}$

QED

5 0

2 years ago