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3 years ago

Solve using any method! 5. x2 + 7x - 8 = 0

Mathematics

2 answers:

Eduardwww [97]3 years ago

8 0

The answer :

86

For x2 + 7x - 8 = 0

Digiron [165]3 years ago

5 0

Answer:

Step-by-step explanation:

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maksim [4K]

Answer: Option 'C' is correct.

Step-by-step explanation:

Since we have given that

Polygon X is similar to polygon.

And Polygon X is three times the size of polygon Y.

So, it becomes,

$\dfrac{\text{Perimeter of polygon X}}{\text{Perimeter of polygon Y}}=\dfrac{3}{1}\\\\\dfrac{81}{x}=\dfrac{3}{1}\\\\x=\dfrac{81}{3}\\\\x=27$

Hence, perimeter of polygon Y is 27.

Therefore, Option 'C' is correct.

5 0

3 years ago

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Solve the inequality -4>-4 + h. A. h<0 B. h>0 C. h<3 D. h>-8

artcher [175]

Answer:

A. H<0

Step-by-step explanation:

Add by 4 from both sides of equation.

-4+4>-4+h+4

Simplify.

0>h

Then, switch sides to find the answer.

h<0

h<0 is the correct answer.

6 0

3 years ago

Each diagram is made of smaller, identical pieces. Tell how many pieces you would shade to model 25%

eduard

Answer:

12 pieces

Step-by-step explanation:

8 0

2 years ago

You want to get from a point A on the straight shore of the beach to a buoy which is 54 meters out in the water from a point B o

anyanavicka [17]

Answer:

$x =\dfrac{45 \sqrt{6}}{ 2}$

Step-by-step explanation:

From the given information:

The diagrammatic interpretation of what the question is all about can be seen in the diagram attached below.

Now, let V(x) be the time needed for the runner to reach the buoy;

∴ We can say that,

$\mathtt{V(x) = \dfrac{70-x}{7}+\dfrac{\sqrt{54^2+x^2}}{5}}$

In order to estimate the point along the shore, x meters from B, the runner should stop running and start swimming if he want to reach the buoy in the least time possible, then we need to differentiate the function of V(x) and relate it to zero.

i.e

The differential of V(x) = V'(x) =0

= $\dfrac{d}{dx}\begin {bmatrix} \dfrac{70-x}{7} + \dfrac{\sqrt{54^2+x^2}}{5} \end {bmatrix}= 0$

$-\dfrac{1}{7}+ \dfrac{1}{5}\times \dfrac{x}{\sqrt{54^2+x^2}}=0$

$\dfrac{1}{5}\times \dfrac{x}{\sqrt{54^2+x^2}}= \dfrac{1}{7}$

$\dfrac{5x}{\sqrt{54^2+x^2}}= \dfrac{1}{7}$

$\dfrac{x}{\sqrt{54^2+x^2}}= \dfrac{1}{\dfrac{7}{5}}$

$\dfrac{x}{\sqrt{54^2+x^2}}= \dfrac{5}{7}$

squaring both sides; we get

$\dfrac{x^2}{54^2+x^2}= \dfrac{5^2}{7^2}$

$\dfrac{x^2}{54^2+x^2}= \dfrac{25}{49}$

By cross multiplying; we get

$49x^2 = 25(54^2+x^2)$

$49x^2 = 25 \times 54^2+ 25x^2$

$49x^2-25x^2 = 25 \times 54^2$

$24x^2 = 25 \times 54^2$

$x^2 = \dfrac{25 \times 54^2}{24}$

$x =\sqrt{ \dfrac{25 \times 54^2}{24}}$

$x =\dfrac{5 \times 54}{\sqrt{24}}$

$x =\dfrac{270}{\sqrt{4 \times 6}}$

$x =\dfrac{45 \times 6}{ 2 \sqrt{ 6}}$

$x =\dfrac{45 \sqrt{6}}{ 2}$

8 0

3 years ago

Help me on this please

insens350 [35]

Answer:

A. x1/15

Step-by-step explanation:

6 0

2 years ago

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