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3 years ago

5

What is the image point of (-7,1) after a translation left 4 units and down 5 units?

1 answer:

irina1246 [14]3 years ago

7 0

The final point should end up at (-11,-4)

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PLEASE HELP ME ANSWER THIS QUESTION! QUICK POINTS!!!

Svetllana [295]

Step-by-step explanation:

2.5<x<8.0???????????

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3 years ago

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The equation of a line in the point-slope form is show below.

Gwar [14]

Answer:

Answer:

your answer is

m= -3

Math-way is a really good site for these kinds of problems i would recommend it if you need anymore help with stuff like this. :)

Step-by-step explanation:

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3 years ago

a 23 gram sample of a substance thats a by-product of fireworks has a K-value of .13. whats the substances half life in days

nata0808 [166]

Given:
Initial mass, m₀ = 23 g
Decay constant, k = 0.13

Exponential decay obeys the equation
$m(t)=m_{0} e^{-kt}$
where t = time, days

For half life, the mass is m = m₀/2.
Therefore
$e^{-0.13t} = \frac{1}{2} \\ -0.13t = ln(0.5)\\t= \frac{ln(0.5)}{-0.13} =5.33$

Answer: The half life is 5.33 days

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2 years ago

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Please solve 5 f <br> (Trigonometric Equations)<br> #salute u if u solved it

Zanzabum

Answer:

$\beta=45\degree\:\:or\:\:\beta=135\degree$

Step-by-step explanation:

We want to solve $\tan \beta \sec \beta=\sqrt{2}$ , where $0\le \beta \le360\degree$ .

We rewrite in terms of sine and cosine.

$\frac{\sin \beta}{\cos \beta} \cdot \frac{1}{\cos \beta} =\sqrt{2}$

$\frac{\sin \beta}{\cos^2\beta}=\sqrt{2}$

Use the Pythagorean identity: $\cos^2\beta=1-\sin^2\beta$ .

$\frac{\sin \beta}{1-\sin^2\beta}=\sqrt{2}$

$\implies \sin \beta=\sqrt{2}(1-\sin^2\beta)$

$\implies \sin \beta=\sqrt{2}-\sqrt{2}\sin^2\beta$

$\implies \sqrt{2}\sin^2\beta+\sin \beta- \sqrt{2}=0$

This is a quadratic equation in $\sin \beta$ .

By the quadratic formula, we have:

$\sin \beta=\frac{-1\pm \sqrt{1^2-4(\sqrt{2})(-\sqrt{2} ) } }{2\cdot \sqrt{2} }$

$\sin \beta=\frac{-1\pm \sqrt{1^2+4(2) } }{2\cdot \sqrt{2} }$

$\sin \beta=\frac{-1\pm \sqrt{9} }{2\cdot \sqrt{2} }$

$\sin \beta=\frac{-1\pm3}{2\cdot \sqrt{2} }$

$\sin \beta=\frac{2}{2\cdot \sqrt{2} }$ or $\sin \beta=\frac{-4}{2\cdot \sqrt{2} }$

$\sin \beta=\frac{1}{\sqrt{2} }$ or $\sin \beta=-\frac{2}{\sqrt{2} }$

$\sin \beta=\frac{\sqrt{2}}{2}$ or $\sin \beta=-\sqrt{2}$

When $\sin \beta=\frac{\sqrt{2}}{2}$ , $\beta=\sin ^{-1}(\frac{\sqrt{2} }{2} )$

$\implies \beta=45\degree\:\:or\:\:\beta=135\degree$ on the interval $0\le \beta \le360\degree$ .

When $\sin \beta=-\sqrt{2}$ , $\beta$ is not defined because $-1\le \sin \beta \le1$

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3 years ago

Can someone help me get y>5x+4

Westkost [7]

Answer:yes

Step-by-step explanation:

8 0

2 years ago