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zheka24 [161]
3 years ago
7

(6x^3 + 18x^2)– 3x^2 =

Mathematics
1 answer:
Debora [2.8K]3 years ago
5 0

https://www.wolframalpha.com/

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Can someone please help me with this question
Pepsi [2]

Answer:

6.9

Explanation:

6 4/5 = 6.8 so 1 decimal point higher than that is 6.9 which is still smaller than 7

4 0
2 years ago
Read 2 more answers
Use the arc length formula to find the length of the curve y = 4x − 5, −1 ≤ x ≤ 3. Check your answer by noting that the curve is
Reptile [31]

Answer:

4\sqrt{17}

Step-by-step explanation:

Let's find the answer by using the arc length formula which is:

\int\limits^a_b {\sqrt{1+(\frac{dy}{dx})^{2} } } \, dx

First, let's find dy/dx which is:

y=4x-5

y'=4*(1)-0

y'=4, now let's use the formula:

\int\limits^3_{-1} {\sqrt{1+4^{2}} } \, dx=\sqrt{17} *(3-(-1))=4\sqrt{17}

Now, using the distance formula we have:

d=\sqrt{(x2-x1)^{2} +(y2-y1)^{2} }

y(-1)=4*(-1)-5=-9 \\y(3)=4*(3)-5=7

So we have two points (-1, -9) and (3, 7) so:

d=\sqrt{(3-(-1))^{2} +(7-(-9))^{2} }=4\sqrt{17}

Notice both equations gave the same length 4\sqrt{17}.

5 0
3 years ago
Which congruence theorem can be used to prove WXS and YZS?
mariarad [96]

I am going to make what is hopefully not an incorrect assumption and say that since the segmentXSZ is black and the other lines are blue, that XS is congruent to ZS. If that be the case we have a side in each triangle that is congruent to each other and an angle that is marked as congruent. Angles XSW and ZSY are vertical angles. By definition, vertical angles are congruent. In each triangle, then, we have an angle, a side and an angle which gives us the congruency postulate ASA.

3 0
3 years ago
((8 ÷ 2 ( 3 + 3 ))) ☆ ☆ ☆ ☆ ☆ ☆ ☆ ☆ ☆ ☆ ☆ ☆ ☆ ☆ ☆ ☆ ☆ ☆ ☆ Answer: ♫・*:.。. .。.:*・ Hewwo frwind! ♫・*:.。. .。.:* I know the answer i
Whitepunk [10]

Answer: it is 24 UWU

Step-by-step explanation:

4 0
3 years ago
Look at the box-and-whisker plot.
Marizza181 [45]

i would say B

when it says "the measure", it's refering to where the second quartile is beginning

3 0
3 years ago
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