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SCORPION-xisa [38]
3 years ago
8

If x = 4, what is the value of 10x + 147

Mathematics
2 answers:
Natali5045456 [20]3 years ago
7 0

Answer:

187

Step-by-step explanation:

10*4=40

40+147=187

NISA [10]3 years ago
3 0

Answer:

251

Step-by-step explanation:

x=4

104 +147=251

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Which x-value is in the domain of the function f (x) = 4cot(2x) + 3?
tigry1 [53]
It’s 24 bc I’m super smart
4 0
2 years ago
What is the product?<br> 5<br> -2<br> -6<br> 2
Nana76 [90]

Answer:

-3

4

Step-by-step explanation:

We do across times down

5*-1 + -2 *-1 = -5+2 = -3

-6 *-1 + 2 * -1 = 6 -2 = 4

8 0
4 years ago
Deb borrows 6,300 for one year at 9.75% how much will she pay back at the end of the loan term
Step2247 [10]

Answer:

<u>The correct answer is that Deb will have to pay US$ 6,914.25 at the end of the loan.</u>

Step-by-step explanation:

<u>Loan in US$:</u> 6,300

<u>Term:</u> One Year

<u>Interest rate: </u>9.75% annually

1. Let's calculate how much money Deb will pay in interests for the loan:

Loan * Interest rate * Term

6,300 * 0.0975 * 1

<u>614,25</u>

<u>Deb will have to pay US$ 614,25 in interests for the loan</u>

2. Let's calculate how much money Deb will pay back at the end of the loan term

Loan in US$ + Interests in US$

6,300 + 614,25

<u>6,914,25</u>

<u>At the end of the loan Deb will have to pay $ 6,914.25</u>

8 0
4 years ago
So I need help with this math question
Dvinal [7]
There is a reflection along the y-axis

so the AB = SG      = AB you may put it if they ask or not

BC =  GP

CA = PS
7 0
4 years ago
Evaluate the integral Find the volume of material remaining in a hemisphere of radius 2 after a cylindrical hole ofradius 1 is d
algol13

You want to find the volume inside the hemisphere x^2+y^2+z^2=4 (i.e. inside the sphere but above the plane z=0) and outside the cylinder x^2+y^2=1. Call this region R.

In cylindrical coordinates, we have

\displaystyle\iiint_R\mathrm dV=\int_0^{2\pi}\int_1^2\int_0^{\sqrt{4-r^2}}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta

\displaystyle=2\pi\int_1^2 r\sqrt{4-r^2}\,\mathrm dr

\displaystyle=-\pi\int_3^0\sqrt u\,\mathrm du

(where u=4-r^2)

\displaystyle=\pi\int_0^3\sqrt u\,\mathrm du

=\dfrac{2\pi}3u^{3/2}\bigg|_0^3=2\sqrt3\,\pi

7 0
4 years ago
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