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Ksju [112]
3 years ago
7

Need help plzz%zzzzzzzzzzzz

Mathematics
1 answer:
SSSSS [86.1K]3 years ago
7 0

Answer:

where is the question plez give question

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marshall27 [118]
87 is the median because 83 and 87's average is 85.
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3 years ago
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To find the product of 42.12 and 10³, move the decimal point in 42.12 __ places to the right because 10³ has __ zeros.
Yakvenalex [24]
<h3>To find the product of 42.12 and 10^3,  move the decimal point in 42.12 3 places to the right because 10^3 has 3 zeros</h3>

<em><u>Solution:</u></em>

Given that,

\text{ product of } 42.12 \text{ and } 10^3

Which means,

42.12 \times 10^3

Here, the exponent of 10 is positive ( which is 3)

When the exponent is positive, we have to move the decimal point to right

When you multiply a number by a power of 10, ( 10!, 10^2, and so on ) move the decimal point of the number to the right the same number of places as the number of zeros in the power of 10

Here, exponent is 3 , therefore move the decimal point right 3 places in 42.12

Therefore,

42.12 \times 10^3 = 42120

7 0
3 years ago
Solve the following system of equations graphically on the set of axes below. y=3x−3 , x+y=5
topjm [15]

Answer:

x=2, y=3

Step-by-step explanation:

First, graph both equations on a rectangular coordinate system (see attached screenshot). Then, simply figure out where these two lines intersect. Since they intersect at (2,3), x=2 and y=3.

Plugging these values in the equations for x and y, you can see if these are the correct answers:

(3)=3(2)-3, (2)+(3)=5

Since both of these are true, you have the right answer.

Hope this helps!

6 0
3 years ago
Which is an advantage of using a credit card? You can make purchases even when you don't have cash. You don't indulge in impulse
mamaluj [8]
You can make purchases even when you don't have cash

hope this helps

8 0
3 years ago
Read 2 more answers
Select the basic integration formula you can use to find the indefinite integral.
kondaur [170]

Answer:

\int \dfrac{du}{u}=\log|u|+C

\int u^n du=\dfrac{u^{n+1}}{n+1}+C

\int \dfrac{du}{u\sqrt{u^2-a^2}}=\dfrac{1}{a}\csc^{-1}(\dfrac{x}{a})+C

Step-by-step explanation:

a.

\int \dfrac{du}{u}=\log|u|+C    [\because \int \dfrac{dx}{x}=\log |x|+C]

b.

\int u^n du=\dfrac{u^{n+1}}{n+1}+C    [\because \int x^n dx=\dfrac{x^{n+1}}{n+1}+C]

c.

\int \dfrac{du}{u\sqrt{u^2-a^2}}=\dfrac{1}{a}\csc^{-1}(\dfrac{x}{a})+C        [\because \dfrac{adx}{x\sqrt{x^2-a^2}}=\csc^{-1}(\dfrac{x}{a})+C]

3 0
3 years ago
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