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2 years ago

Is 10.2 greater or lesser than 10

Mathematics

2 answers:

Westkost [7]2 years ago

8 0

10.2 is greater than 10.

insens350 [35]2 years ago

3 0

Its is greater, your welcome xoxo savi eden

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PLEASEEE HELP MEEE!! I DONT UNDERSTAND AT ALL!!

jenyasd209 [6]

Domain: (-∞,∞)

Range: (3,∞)

x-intercepts: none

y-intercepts: (0,7)

Interval positive: (3,∞)

Interval negative: none

Interval increasing: (7,∞)

Interval decreasing: (-∞,7)

I'm not sure what the average rate of change over is though.

7 0

2 years ago

100, 91, 82, 73 Number pattern expression equations

emmasim [6.3K]

Answer:

The pattern is subtract 9 (OR -9)

Step-by-step explanation:

100 - 91 = 9

91 - 82 = 9

82 - 73 = 9

Glad I could help! Feel free to mark brainliest!

5 0

2 years ago

Read 2 more answers

Show <br> a way to count from $82 to $512

gogolik [260]

You could add two until you get to 512

5 0

2 years ago

The radius of a circular plate is 12 centimeters. What is the area, in square centimeters, of this plate? Write your Anwer using

lesya [120]

Answer:

The area of circular plate is <u>144π square centimeters</u>.

Step-by-step explanation :

<h3><u>Solution</u> :</h3>

Here, we have given that the radius of circular plate is 12 centimeters.

So, finding the area of circular plate by substituting the values in the formula :

$\longrightarrow{\pmb{\sf{Area_{(Circle)} = \pi{r}^{2}}}}$

$\longrightarrow{\sf{Area_{(Circle)} = \pi{(12)}^{2}}}$

$\longrightarrow{\sf{Area_{(Circle)} = \pi{(12 \times 12)}}}$

$\longrightarrow{\sf{Area_{(Circle)} = \pi{(144)}}}$

$\longrightarrow{\sf{Area_{(Circle)} = \pi \times 144}}$

$\longrightarrow{\sf{Area_{(Circle)} = 144\pi}}$

$\star{\underline{\boxed{\frak{\purple{Area_{(Circle)} = 144\pi\: {cm}^{2}}}}}}$

Hence, the area of circular plate is 144π square centimeters.

$\rule{300}{2.5}$

5 0

2 years ago

Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.

seropon [69]

Answer:

$h'(x)=\frac{3r^{2}}{2\sqrt{r^3+5}}$

Step-by-step explanation:

1) The Fundamental Theorem of Calculus in its first part, shows us a reciprocal relationship between Derivatives and Integration

$g(x)=\int_{a}^{x}f(t)dt \:\:a\leqslant x\leqslant b$

2) In this case, we'll need to find the derivative applying the chain rule. As it follows:

$h(x)=\int_{a}^{x^{2}}\sqrt{5+r^{3}}\therefore h'(x)=\frac{\mathrm{d} }{\mathrm{d} x}\left (\int_{a}^{x^{2}}\sqrt{5+r^{3}}\right )\\h'(x)=\sqrt{5+r^{3}}\\Chain\:Rule:\\F'(x)=f'(g(x))*g'(x)\\h'=\sqrt{5+r^{3}}\Rightarrow h'(x)=\frac{1}{2}*(r^{3}+5)^{-\frac{1}{2}}*(3r^{2}+0)\Rightarrow h'(x)=\frac{3r^{2}}{2\sqrt{r^3+5}}$

3) To test it, just integrate:

$\int \frac{3r^{2}}{2\sqrt{r^3+5}}dr=\sqrt{r^{3}+5}+C$

5 0

2 years ago