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2 years ago

Use the figure to find the measures of the numbered angles.

Mathematics

1 answer:

Elena L [17]2 years ago

5 0

Step-by-step explanation:

∠5 = 49⁰

∠6 = 131⁰

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8. The high school graduating class of 2000 was 260 students. Five years later, the graduating class was 320 students. By 2010,

olga nikolaevna [1]

Answer:

the question is confusing but the answer is 2010

Step-by-step explanation:

because in the year 2000 it was

260

then increased by

over 5 years making the next class

320

then in 2010 it was

400

and 320 + what is 400?

so the answer is:

2010

4 0

2 years ago

A grocery store sells a bag of 6 oranges for $4.20. If Ayden spent $2.80 on oranges, how many did he buy?

aleksley [76]

Answer:

4 oranges.

Step-by-step explanation:

First, you find the unit rate of cost per orange, which is $0.70, then divide that by $2.80.

5 0

3 years ago

I need help what is 1/3x-3?

lilavasa [31]

Answer:

$\frac{1}{3}x-3=\frac{x}{3}-3$

Step-by-step explanation:

$\frac{1}{3}x-3\\\\\frac{1}{3}x = \frac{x}{3} \\\\Find\:L.C.M.\\\\=\frac{x}{3}-3$

8 0

2 years ago

Find the maximum value or minimum value for the function f(x) = 0.15(x + 1)² - 3.

il63 [147K]

Answer:

The minimum value for $f(x) = 0.15(x + 1)^2 - 3$ is $-3$ .

Step-by-step explanation:

Given function is $f(x) = 0.15(x + 1)^2 - 3$

We need to find the maximum value or the minimum value for the function.

Now, differentiate $f(x) = 0.15(x + 1)^2 - 3$ w.r.t $x$ .

$f'(x) =\frac{d}{dx}(0.15(x + 1)^2 - 3)\\f'(x)=\frac{d}{dx}(0.15(x+1)^2-\frac{d}{dx}(3)\\$

$f'(x)=2\times 0.15(x+1)\frac{d}{dx}(x+1)-0\\f'(x)=0.3(x+1)(1)\\f'(x)=0.3(x+1)$

Now, we will equate $f'(x)=0$ to find critical point.

$0.3(x+1)=0\\x=-1$

Plug this critical point in to the function $f(x) = 0.15(x + 1)^2 - 3$ we get,

$f(-1) = 0.15(-1 + 1)^2 - 3\\f(-1)=-3$

Also, $f''(x)=0.3$ which is positive, We have minimum value.

So, the minimum value for $f(x) = 0.15(x + 1)^2 - 3$ is $-3$ .

7 0

2 years ago

a crowbar of 2cm is used to lift an object if 800N. if the effort arm is 260cm. calculate the effort applied.

Gelneren [198K]

Answer:

solution:Here

Given,

Load=800 N

Load distance=2cm

Effort distance=260cm

Effort applied=?

We know that,

Load × Load distance = Effort × Effort distance

or, 800×2 = e × 260

or,1600 = 260e

or,e = 1600/260

Thus, e = 6.15 N

Hence, The effort applied = 6.15 N

Hope this answer helped you..

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8 0

2 months ago