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3 years ago

5

Algebra help plss222

2 answers:

Keith_Richards [23]3 years ago

6 0

Answer:

y=1/9

Step-by-step explanation:

nadya68 [22]3 years ago

3 0

Answer:

I'm a little unsure but I think it should be y = $-\frac{1}{9}$

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A tree 63 ft high casts a 38 ft shadow. How tall is a tree that casts a 52 ft shadow at the same time of the day? (to the neares

Mrac [35]

52/38 = 1.368

1.368 x 63 = 86.21

tree is 86.2 feet high

3 0

3 years ago

Polynormual sum of (x^2+9)+(-3x^2-11x+4)

amid [387]

Answer:

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Step-by-step explanation:

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5 0

3 years ago

Simplify.<br> (-3.9) + (6.01) - (-7.423)<br> -5.513.<br> 2.487.<br> 9.533.

lianna [129]

First, since we must follow PEMDAS, we need to do the operation that is first, which is addition.

-3.9 + 6.01 ← Adding a negative means subtracting, then giving the sign of the bigger number.

So now we solved those and we got this answer: 2.11

Now, we need to subtract and re-write the expression.

2.11 - (-7.423) = 2.11 + 7.423= 9.533

Ok....so now we have our final answer, and we have the choices above.

The only answer that matches ours is the last one.

Final answer: 9.533

Hope I helped ^_^

4 0

3 years ago

Read 2 more answers

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 $-1\le \sin \beta \le1$

4 0

3 years ago

What’s is the vertex y=x^2-2x+26

andreev551 [17]

Answer:

(1, 25)

Step-by-step explanation:

Convert to vertex form so that the equation is (x - 1)^2 + 25

If 1 is <em>h </em>and 25 is <em>k</em>, (<em>h, k</em>), the result would be (1, 25)

5 0

3 years ago