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3 years ago

Jake rounded 18,752 to the nearest ten thousand which is his number

Mathematics

2 answers:

cluponka [151]3 years ago

7 0

20,000 is the Answer. Hope this helps.

Effectus [21]3 years ago

4 0

The nearest ten thousand is 20,000

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Show calculations for each of the ratio table

Sladkaya [172]

Answer:

From left to right:

3, 30, 20

Step-by-step explanation:

Start by determining the ratio of yellow to red paint. Use the column where both amounts are given:

yellow to red = 2 to 6 = 1 to 3

2 to 6 the same as 1 to 3 when reduced, by dividing both sides by 2.

This means yellow paint is a third of red paint. Backwards, red paint is three times yellow paint.

To go from yellow to red, do yellow X 3.

To go from red to yellow, do red ÷ 3.

Column 1:

1 X 3 = 3

Column 3:

10 X 3 = 30

Column 4:

60 ÷ 3 = 20

The ratios are:

1 to 3

2 to 6

10 to 30

20 to 60

4 0

3 years ago

Can someone help with this question

NikAS [45]

1st : Find the volume of the rectangular block:
12 × 10 × 20 = 2400 cm^3.

2nd : Find the volume of a cylinder
$\pi {r}^{2} h$
Where r is radius & h is height.
so:
$\pi \times 2.2 {}^{2} \times 5.0$
= 76.02654222 cm^3

Lastly, divide the volume of the rectangular block by the volume of the cylinder
2400 ÷ 76.026 = 31.5679226

You can't round it up so the answer is F) 31

4 0

3 years ago

Solve using algebraic equation: <br> 5sin2x=3cosx<br><br> (No exponents)

DaniilM [7]

$5\sin2x=3\cos x\iff10\sin x\cos x=3\cos x$

by the double angle identity for sine. Move everything to one side and factor out the cosine term.

$10\sin x\cos x=3\cos x\iff10\sin x\cos x-3\cos x=\cos x(10\sin x-3)=0$

Now the zero product property tells us that there are two cases where this is true,

$\begin{cases}\cos x=0\\10\sin x-3=0\end{cases}$

In the first equation, cosine becomes zero whenever its argument is an odd integer multiple of $\dfrac\pi2$ , so $x=\dfrac{(2n+1)\pi}2$ where $n[/tex ]is any integer.\\Meanwhile,\\[tex]10\sin x-3=0\implies\sin x=\dfrac3{10}$

which occurs twice in the interval $[0,2\pi)$ for $x=\arcsin\dfrac3{10}$ and $x=\pi-\arcsin\dfrac3{10}$ . More generally, if you think of $x$ as a point on the unit circle, this occurs whenever $x$ also completes a full revolution about the origin. This means for any integer $n$ , the general solution in this case would be $x=\arcsin\dfrac3{10}+2n\pi$ and $x=\pi-\arcsin\dfrac3{10}+2n\pi$ .

6 0

3 years ago

URGENT HELP PLS 20 POINTS

horrorfan [7]

Answer:

I think i knew it but I forget

sorry!

Step-by-step explanation:

7 0

2 years ago

Given f(x) = 10 - 2x, find f(7).

Dmitry_Shevchenko [17]

Answer:

-4

Step-by-step explanation:

f(x) = 10 - 2x,

Let x= 7

f(7) = 10 2(7)

= 10-14

= -4

7 0

3 years ago