All categories

3 years ago

Find the perimeter and area of each figure below.

Mathematics

1 answer:

KonstantinChe [14]3 years ago

6 0

A) Perimeter of Trapezium = 8+5+6+15.5
= 34.5in
Area= 1/2*(a+b)*h
=1/2*(8+15.5)*4
=47in^2

B)Perimeter= 12+6+8.5
=26.5mm

Area=1/2*b*h
=1/2*12*4
=24mm^2

You might be interested in

SOMEONE HELP ME PLSSSSSSSSSSSSSSSSSSS

fomenos

Answer: drag the first one to the fourth one, drag the second one to the first one,drag the third one to the second one,drag the fourth one to the third one.

Step-by-step explanation:if its wrong then im sorry

3 0

3 years ago

5x^(-2)-17x^(-1)+6<br> i’m so confused pls help<br> will give brainliest

laila [671]

It's a quadratic equation in disguise. If you let $y=x^{-1}$ , then $y^2=x^{-2}$ , and we can rewrite the equation as

$5y^2-17y+6=0$

Solve for $y$ however you like; we get $y=\dfrac25$ and $y=3$ .

But we want to solve for $x$ , so we have

$y=x^{-1}=\dfrac25\implies x=\dfrac52$

$y=x^{-1}=3\implies x=\dfrac13$

5 0

3 years ago

In Exercises 40-43, for what value(s) of k, if any, will the systems have (a) no solution, (b) a unique solution, and (c) infini

svet-max [94.6K]

Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

The system cannot have unique solution.

Step-by-step explanation:

We have the following system of equations

$x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2$

The augmented matrix is

$\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]$

The reduction of this matrix to row-echelon form is outlined below.

$R_2\rightarrow R_2-R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]$

$R_3\rightarrow R_3-2R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]$

$R_3\rightarrow R_3-R_2$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]$

The last row determines, if there are solutions or not. To be consistent, we must have k such that

$k^2-k-2=0$

$\left(k+1\right)\left(k-2\right)=0\\k=-1,\:k=2$

Case k = −1:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]$

If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]$

This gives the infinite many solution.

5 0

3 years ago

576 = 96 × 6

aliya0001 [1]

Answer:

Step-by-step explanation

A.96=6+576

B.576=96+6

C. 96=6×576

D. 576=96×6

5 0

3 years ago

A intercept of 2 and slope of 1/6

Katen [24]

Answer:

what is your question though. if its create an equation, its

y = 1/6x + 2

3 0

3 years ago