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4 years ago

Find an expression for the area of this rectangle (2x+1) (x+8)

2 answers:

8 0

The correct awnser is this is not a perfect rectangle because it does not have a polynomial value I hope I helped you out with this problem if so please feel free to mark me the brainliest or to give me a thanks

stepan [7]4 years ago

4 0

Assuming the brackets you have given are two of the sides of the rectangle (I apologise if that's not the case), you just need to multiply the brackets together, which means multiplying each term. Therefore: 2x × x, 2x × 8, 1 × x, and 1 × 8. The answers to these are 2x^2, 16x, x, and 8. Then you just add these terms to get 2x^2 + 17x + 8

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Quadrilateral EFGH is inscribed inside a circle as shown below. Write a proof showing that angles H and F are supplementary. Cir

vladimir1956 [14]

Answer:

That is called a cyclic quadrilateral. Angles E and G add up to 180 degrees and angles H and F add up to 180 degrees.

Supplementary angles are two angles whose measures add up to 180°

7 0

3 years ago

Need help please will mark brainliest

Bumek [7]

$2x + 4x = 180 \\ 6x = 180 \\ x = 30$

8 0

3 years ago

Read 2 more answers

Is the 161 rounded to the nearest 10 or to the nearest 100

kaheart [24]

Answer:

Step-by-step explanation:

so you can make it 200

5 0

4 years ago

Please answer ASAP, I'm on the clock, please!

klio [65]

<span>Student 2: -9 + (6 + 9) - 42
-9 + (9 + 6) - 42
(-9 + 9) + 6 - 42
0 + 6 - 42
0 + 6 - 16
0 + 6 + (-16)
6 + (-16)
-10</span>

student 2 did

hope this helps

since it is only addition and subtraction, you can mix the orders up, as long as you use the correct sign

8 0

3 years ago

AP Calculus Help Please!

ivanzaharov [21]

4. By the fundamental theorem of calculus,

$\displaystyle\frac{\mathrm dg(x)}{\mathrm dx}=\frac{\mathrm d}{\mathrm dx}\int_{-2}^xf(t)\,\mathrm dt=f(x)$

$g(x)$ is increasing on those intervals where $g'(x)=\dfrac{\mathrm dg(x)}{\mathrm dx}>0$ . So you have

$g'(x)=f(x)=\begin{cases}3&\text{for }-3\le x$

which is clearly positive for $x\in[-3,0)$ , and in the second interval you have

$-x+3>0\implies x$

Together, this means $g'(x)>0$ for all $x\in[-3,3)$ .

5. When $0\le x\le6$ , $f(x)$ reduces to $-x+3$ , so you have

$g(x)=\displaystyle\int_{-2}^xf(t)\,\mathrm dt=\int_{-2}^03\,\mathrm dt+\int_0^x(-t+3)\,\mathrm dt$
$g(x)=6+\left(3t-\dfrac12t^2\right)\bigg|_{t=0}^{t=x}$
$g(x)=6+3x-\dfrac12x^2$

4 0

4 years ago