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8_murik_8 [283]

3 years ago

14

Numbers to the nearest ten 48393 1549 37,519 65,810 2175

2 answers:

lidiya [134]3 years ago

5 0

Answer:

6787878

Step-by-step explanation:

katen-ka-za [31]3 years ago

4 0

Answer:

48390,"37,520","65,810",2170 or 2180 (both will work)

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HELP ASAP!!<br>-4y + 7 + 9y - 3

gulaghasi [49]

Answer:

5y+4

Step-by-step explanation:

Combine Like Terms:

=−4y+7+9y+−3

=(−4y+9y)+(7+−3)

=5y+4

5 0

3 years ago

Read 2 more answers

Joy has ₱500.00 to spend for food. She spent ₱98.75 on Monday, ₱73.75 on Tuesday and ₱50.25 on Wednesday. How much money was lef

Alex787 [66]

Answer:

₱277.25

Step-by-step explanation:

₱500.00 - ₱98.75 - ₱73.75 - ₱50.25 = ₱277.25

8 0

3 years ago

The Empire State Building weighs about 7.3×108pounds. The One World Trade Center building weighs about 88,200,000 pounds. What i

In-s [12.5K]

Answer:

$8.182X10^8 $pounds$

a=8.182 and b=8

Step-by-step explanation:

Weight of the Empire State Building = $7.3X 10^8$ pounds.

Weight of the One World Trade Center building= 88,200,000 pounds.

= $8.82 X 10^7$

The addition of the two:

$=7.3X 10^8+8.82 X 10^7\\$To make it easier to add, express both as powers of 8\\=7.3X 10^8+0.882 X 10^8\\=(7.3+0.882)X10^8\\=8.182X10^8 $ pounds$

Comparing with the form: $aX10^b$

a=8.182 and b=8

4 0

3 years ago

If the family's net monthly income is $7,800, what percent of the income is spent on food, clothing, and housing?

Gala2k [10]

Answer:

60 percent

Step-by-step explanation:

5 0

3 years ago

Solve the initial value problem y'=2 cos 2x/(3+2y),y(0)=−1 and determine where the solution attains its maximum value.

zloy xaker [14]

Answer:

$y=\frac{-3\pm\sqrt{4sin2x+1}}{2}$

$x={\pi}{4}$

Step-by-step explanation:

We are given that

$y'=\frac{2cos2x}{3+2y}$

y(0)=-1

$\frac{dy}{dx}=\frac{2cos2x}{3+2y}$

$(3+2y)dy=2cos2x dx$

Taking integration on both sides then we get

$\int (3+2y)dy=2\int cos 2xdx$

$3y+y^2=sin2x+C$

Using formula

$\int x^n=\frac{x^{n+1}}{n+1}+C$

$\int cosx dx=sinx$

Substitute x=0 and y=-1

$-3+1=sin0+C$

$-2=C$

$sin0=0$

Substitute the value of C

$y^2+3y=sin2x-2$

$y^2+3y-sin 2x+2=0$

$y=\frac{-3\pm\sqrt{(3)^2-4(1)(-sin2x+2)}}{2}$

By using quadratic formula

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$y=\frac{-3\pm\sqrt{9+4sin2x-8}}{2}=\frac{-3\pm\sqrt{4sin2x+1}}{2}$

Hence, the solution $y=\frac{-3\pm\sqrt{4sin2x+1}}{2}$

When the solution is maximum then y'=0

$\frac{2cos2x}{3+2y}=0$

$2cos2x=0$

$cos2x=0$

$cos2x=cos\frac{\pi}{2}$

$cos\frac{\pi}{2}=0$

$2x=\frac{\pi}{2}$

$x=\frac{\pi}{4}$

3 0

3 years ago