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3 years ago

HELP ME PLEASE I REALLY NEED HELP!? :(

Mathematics

1 answer:

Aleksandr [31]3 years ago

5 0

Just answers since long and many simplifiction
it's like doing those multiplication memory things that had like 100 questions but now with algebra fractions

5. fourth or $\frac{5x-15}{x^{2}-3x-28}$
6. third or $\frac{4x+25}{x^{2}+5x+4}$
7.the answe ris x+3 since it can be factored out of the 5x+15 tingie (second )
8. fourth or $\frac{x^{2}+4x}{4}$
9. fourth or $\frac{2y+6}{9y-7}$ times $\frac{3}{2}$
10. third or $\frac{(x+3)(x+8)}{(x-8)(x-8)}$
11. second or $\frac{x-5}{x-4}$
12. third or 11(x+7)=60

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Answer: 5.22 seconds

<u>Step-by-step explanation:</u>

t represents time and y represents the height.

Since we want to know when the ball hits the ground, find t when y = 0

Ball 1 starts at a height of 109 --> h = 109

0 = -16t² + 109

16t² = 109

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$t=\sqrt{\dfrac{109}{16}}$

$t=\dfrac{\sqrt{109}}{2}$

t = 5.22

4 0

3 years ago

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3 years ago

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Basile [38]

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8 0

3 years ago

Right triangle ABC is shown. Triangle A B C is shown. Angle A C B is a right angle and angle C B A is 50 degrees. The length of

anzhelika [568]

Question:

Obtian the equation that can be used to solve for c

Answer:

Sin50 = 3/c

Step-by-step explanation:

A triangle showing the information given has been attached below for clarity :

Using trigonometry :

AC = 3 = opposite side of angle 50°

AB = c = hypotenus

Using the sine relation of a right angled triangle :

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6 0

3 years ago

How do you do these?

Assoli18 [71]

$\bf 6\sqrt[4]{405}\quad now\to 
\begin{cases}
405\to 3\cdot 3\cdot 3\cdot 3\cdot 5\\
\qquad 3^4\cdot 5
\end{cases}
\\ \quad \\
6\sqrt[4]{3^4\cdot 5}\implies 6\cdot 3\sqrt[4]{5}\\
----------------------------\\
$ $\bf 3\sqrt{49x^9y^{16}}\quad now\to 
\begin{cases}
49\to 7\cdot 7\to 7^2\\
x^9\to x^{2+2+2+2+1}\\
\qquad x^2\cdot x^2\cdot x^2\cdot x^2\cdot x^1\\
\qquad (x^2)^4\cdot x\\
y^{16}\to y^{2+2+2+2+2+2+2+2}\\
\qquad y^2\cdot y^2\cdot y^2\cdot y^2\\
\qquad \cdot y^2\cdot y^2\cdot y^2\cdot y^2\\
\qquad (y^2)^8
\end{cases}
\\ \quad \\
3\sqrt{7^2(x^2)^4x(y^2)^8}\implies 3\sqrt{7^2(x^4)^2x(y^8)^2}
\\ \quad \\
3\cdot 7\cdot x^4\cdot y^8\sqrt{x}\implies 21x^4y^8\sqrt{x}$

5 0

3 years ago