All categories

3 years ago

Given: △ABC≅△DEF, △KLM≅△DEF

Mathematics

1 answer:

lesantik [10]3 years ago

6 0

I reccomend that you add an image or better explain what you want us to do.

You might be interested in

5/8 -0.5 evaluate each expression. Write answer as a decimal

gregori [183]

Answer:

0.125

Step-by-step explanation:

0.5 =1/2

Now

5/8 - 1/2

5/8 - 1*4/2*4

5/8-4/8

(5-4)/8

1/8

0.125

6 0

3 years ago

Martha is making a scarf for her sister each day she knits 1/6 of a scarf what fraction of the scarf will be complete after thre

schepotkina [342]

Multiply 1/6 by 3 =1/2
In 3 days she would have completed 1/2 of the scarf.

8 0

3 years ago

Read 2 more answers

Use 3.14 for the value of , and round your answers to two decimal places, where applicable. If the diameter of a circle is 15 m,

sergejj [24]

if the diameter of a circle is 15, its radius is half that or 7.5.

$\bf \textit{area of a circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=7.5 \end{cases} A=\pi (7.5)^2\implies A=56.25\pi \implies \stackrel{\pi =3.14}{A=176.625}$

7 0

3 years ago

Use the fundamental theorem of calculus to find the area of the region between the graph of the function x^5 + 8x^4 + 2x^2 + 5x

BaLLatris [955]

Answer:

The area of the region is 25,351 $units^2$ .

Step-by-step explanation:

The Fundamental Theorem of Calculus: if $f$ is a continuous function on $[a,b]$ , then

$\int_{a}^{b} f(x)dx = F(b) - F(a) = F(x) | {_a^b}$

where $F$ is an antiderivative of $f$ .

A function $F$ is an antiderivative of the function $f$ if

$F^{'}(x)=f(x)$

The theorem relates differential and integral calculus, and tells us how we can find the area under a curve using antidifferentiation.

To find the area of the region between the graph of the function $x^5 + 8x^4 + 2x^2 + 5x + 15$ and the x-axis on the interval [-6, 6] you must:

Apply the Fundamental Theorem of Calculus

$\int _{-6}^6(x^5+8x^4+2x^2+5x+15)dx$

$\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\\\\int _{-6}^6x^5dx+\int _{-6}^68x^4dx+\int _{-6}^62x^2dx+\int _{-6}^65xdx+\int _{-6}^615dx$

$\int _{-6}^6x^5dx=0\\\\\int _{-6}^68x^4dx=\frac{124416}{5}\\\\\int _{-6}^62x^2dx=288\\\\\int _{-6}^65xdx=0\\\\\int _{-6}^615dx=180\\\\0+\frac{124416}{5}+288+0+18\\\\\frac{126756}{5}\approx 25351.2$

3 0

3 years ago

I don't really understand this problem

yKpoI14uk [10]

Complementary angles add up to 90 (there should be a right angle in the diagram)

x+20+x+10 needs to equal 90 where x=30

Left angle is measured 30+20=50 degrees

Angle on the right is measured 30+10=40 degrees

4 0

3 years ago