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3 years ago

A triangle with a base of 9 meters has an altitude of 12 meters. Which would give the area of the triangle?

Mathematics

2 answers:

castortr0y [4]3 years ago

5 0

StepAnswer:

i think b

-by-step explanation:

to find the area of the triangle its base times height divided by 2

Ierofanga [76]3 years ago

4 0

the correct answer is B 1/2 x 9 x 12 = 54

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Given: m∠LOI=38° m∠JOP=84° Find: m∠JKP

grigory [225]

Answer:

84°

Step-by-step explanation:

I think your question is missed of key information, allow me to add in and hope it will fit the original one. Please have a look at the attached photo.

Given that:

∠LOI=38°
∠JOP=84 °

As we can see from the figure that ∠JOP and ∠JKP are on the same segment JP. Hence, ∠JOP and ∠JKP are congruent and the angle of ∠JKP is 84°

8 0

3 years ago

Given the differential Equation y'+2y=2e^x ;solve this equation using the integration factor; solve for y to get the general sol

SCORPION-xisa [38]

$y'+2y=2e^x\Longrightarrow y'=2e^x-2y$

If $f'(x)=g(x)$ then $y=\int{g(x)dx}$

So we extract,

$y=\int{2e^x-2x}dx$

Which becomes,

$y=2e^x-x^2+C$

Hope this helps.

r3t40

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3 years ago

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ivann1987 [24]

Answer:the answer will be 6

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Triangle ABC is to be dilated through point P with a scale factor of 3. How many units away from point A along ray PA will A’ be

drek231 [11]

Answer:

A' will be located 10 units from point A along ray PA

Step-by-step explanation:

we know that

The scale factor is equal to 3

To obtain PA', multiply PA by the scale factor

PA'=PA*3

PA=5 units

substitute

PA'=(5)*3=15 units

AA'=PA'-PA=15-5=10 units

therefore

A' will be located 10 units from point A along ray PA

3 0

3 years ago

Read 2 more answers

Solve 73 make sure to also define the limits in the parts a and b

Aleks04 [339]

73.

$f(x)=\frac{3x^4+3x^3-36x^2}{x^4-25x^2+144}$

$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3$ $\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3\cdot\frac{1}{2}=3$

Since we can't divide by zero, we need to find when:

$x^4-2x^2+144=0$

But before, we can factor the numerator and the denominator:

$\begin{gathered} \frac{3x^2(x^2+x-12)}{x^4-25x^2+144}=\frac{3x^2((x+4)(x-3))}{(x-3)(x-3)(x+4)(x+4)} \\ so: \\ \frac{3x^2}{(x+3)(x-4)} \end{gathered}$

Now, we can conclude that the vertical asymptotes are located at:

$\begin{gathered} (x+3)(x-4)=0 \\ so: \\ x=-3 \\ x=4 \end{gathered}$

so, for x = -3:

$\lim_{x\to-3^-}f(x)=\lim_{x\to-3^-}-\frac{162}{x^4-25x^2+144}=-162(-\infty)=\infty$ $\lim_{x\to-3^+}f(x)=\lim_{x\to-3^+}-\frac{162}{x^4-25x^2+144}=-162(\infty)=-\infty$

For x = 4:

$\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty$ $\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty$

4 0

1 year ago