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3 years ago

7 cm 4 cm 8 cm what is the area?

Mathematics

2 answers:

adell [148]3 years ago

5 0

what shape is this

I cant answer it if there's no shape in the question

Lena [83]3 years ago

3 0

7cm

4cm

8cm

Aare=?

soln

a+b+c

7cm+4cm+8cm

A=19cm

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jeff hiked for 2 hours and traveled 5 miles. If he continues at the same pace, which equation will show the relationship between

AfilCa [17]

His rate = d/t because d=rt
So r = 5/2 = 2.5 mi/hr
So d = 5/2•t

5 0

3 years ago

Read 2 more answers

The sum of all but one interior angle of a heptagon is 776Â°. What is the measure of the final interior angle? 42Âº 56Âº 124Âº 3

weqwewe [10]

The measure of the seventh interior angle of the heptagon is 124°. (Correct choice: C)

<h3>What is the measure of the missing interior angle in a heptagon?</h3>

Heptagons are polygons with seven sides, seven vertices, seven interior angles and seven central angles. Herein we know the value of the sum of six interior angles and we need to know the measure of the seventh interior angle. We can determine the measure of the seven interior angles by using the following expression:

θ = (n - 2) · 180°, where n is the number of sides of the polygon. (1)

If we know that n = 7, then sum of the internal angles in the heptagon is:

θ = (7 - 2) · 180°

θ = 900°

And the measure of the final interior angle is found by subtraction:

θ₇ = 900° - 776°

θ₇ = 124°

The measure of the seventh interior angle of the heptagon is 124°. (Correct choice: C)

To learn more on polygons: brainly.com/question/17756657

#SPJ1

3 0

2 years ago

The coordinates of the vertices of triangle ABC are A(3,6), B(6, 3), and C(9, 9). If triangle ABC Is dilated by 1/3x and 1/3y to

katovenus [111]

Answer:

$A' = (1,2)$

$B'=(2,1)$

$C' =(3,3)$

Step-by-step explanation:

Given

$A = (3,6)$

$B = (6,3)$

$C = (9,9)$

$Scale = \frac{1}{3}$

Required

Determine the vertices of ABC

If truly the question asks for the vertices of ABC, then the vertices are:

$A = (3,6)$ $B = (6,3)$ $C = (9,9)$

However, I'm quite sure that's not the requirement of the question. So, I'll solve for A'B'C'

To do this, we simply multiply the vertices of ABC by the scale of dilation.

$(A'B'C') = Scale\ Factor * (ABC)$

$A' = \frac{1}{3} * A$

$A' = \frac{1}{3} * (3,6)$

$A' = (\frac{3}{3},\frac{6}{3})$

$A' = (1,2)$

$B' = \frac{1}{3} * B$

$B' = \frac{1}{3} * (6,3)$

$B' = (\frac{6}{3},\frac{3}{3})$

$B'=(2,1)$

$C' = \frac{1}{3} * C$

$C' = \frac{1}{3} * (9,9)$

$C' = (\frac{9}{3},\frac{9}{3})$

$C' =(3,3)$

6 0

3 years ago

Differentiating a Logarithmic Function In Exercise, find the derivative of the function.

UkoKoshka [18]

Answer:

$\frac{d}{dx} ln(\frac{x(x-1)}{x-2}) = \frac{x-2}{x(x-1)} \frac{x^2 -4x +2}{(x-2)^2} = \frac{x^2 -4x +2}{x(x-1)(x-2)}$

Step-by-step explanation:

For this case we want to find the derivate of this function:

$y = ln(\frac{x(x-1)}{x-2})$

And in order to find the derivate we need to apply the chain rule given by:

$\frac{df(u)}{dx} =\frac{df}{du} \frac{du}{dx}$

And on this case $f = ln(u), u = \frac{x(x-1)}{x-2}$

And we can find the partial derivates like this:

$\frac{d}{du} (ln(u))= \frac{1}{u}$

$\frac{d}{dx}(\frac{x(x-1)}{x-2})= \frac{(2x-1)(x-2) -x(x-1)}{(x-2)^2}= \frac{x^2 -4x +2}{(x-2)^2}$

And if we replace we got:

$\frac{d}{dx} ln(\frac{x(x-1)}{x-2}) = \frac{1}{u} \frac{x^2 -4x +2}{(x-2)^2}$

And if we replace $u = \frac{x(x-1)}{x-2}$ we got:

$\frac{d}{dx} ln(\frac{x(x-1)}{x-2}) = \frac{x-2}{x(x-1)} \frac{x^2 -4x +2}{(x-2)^2} = \frac{x^2 -4x +2}{x(x-1)(x-2)}$

And that would be our final answer on this case

7 0

4 years ago

Help me ineed a good grade becasue i need a good grade

yulyashka [42]

ANSWER I know one is 2/6

6 0

3 years ago