3 years ago

Are the triangles below similar?

Mathematics

1 answer:

HACTEHA [7]3 years ago

8 0

Answer:

I think it's C

You might be interested in

If a=-6 and b=-10 evaluate the following: a + 6

ser-zykov [4K]

Answer:

-6+-10=-16

Step-by-step explanation:

+ - = -

........

6 0

3 years ago

Estimate the measure of the clockwise angle formed by the minute hand and the second hand, then classify the angle.

irga5000 [103]

A. 54°; acute is the answer

8 0

4 years ago

Is (x+7) a factor of f(x)= x^3-3x^2+2x-8

nikklg [1K]

Answer:

Step-by-step explanation:

Use synthetic division to answer this. If the remainder is zero, then we can safely assume the divisor (x + 7) is a factor of the polynomial f(x)= x^3-3x^2+2x-8.

We use -7 as the divisor in synth. div. This comes from the factor (x + 7):

-7 / 1 3 2 -8

-7 28 -210

-------------------------

1 -4 30 -218

Here, the remainder is -218, not zero, so no, (x+7) is not a factor of f(x)= x^3-3x^2+2x-8.

6 0

4 years ago

34R x R = 1725 what kind of problem is this

mel-nik [20]

Answer:

i dont know it look stupid

Step-by-step explanation:

5 0

3 years ago

[(1/(3+x))-(1/3)] / x Limit 0

Effectus [21]

First find a common denominator and combine the fractions in the numerator:

$\displaystyle\lim_{x\to0}\frac{\dfrac1{3+x}-\dfrac13}x=\lim_{x\to0}\frac{\dfrac3{3(3+x)}-\dfrac{3+x}{3(3+x)}}x=\lim_{x\to0}\frac{3-(3+x)}{3x(3+x)}$

Now simplify and cancel out all the terms that you can:

$\displaystyle\lim_{x\to0}\frac{3-3-x}{3x(3+x)}=-\frac13\lim_{x\to0}\frac1{3+x}$

Since the remaining expression is continuous as a function of $x$ , you can directly substitute to end up with

$\displaystyle-\frac13\lim_{x\to0}\frac1{3+x}=-\frac13\times\frac1{3+0}=-\frac19$

8 0

3 years ago