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3 years ago

Simplify (-4x^2-3y-8)+(5y-2x^2+1)

Mathematics

1 answer:

ddd [48]3 years ago

8 0

Add the like terms.

The like terms in this case are:

-4x^2 and -2x^2

This makes -6x^2

-3y and 5y

This makes 2y

-8 and 1

This makes -7

Your answer is (-6x^2+2y-7)

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IceJOKER [234]

(3,4) and (-5,6) are "coordinate planes".

These appear in algebra and math when you're graphing. These coordinate planes consist of "x" and "y" (x,y). The x's (which are 3 and -5 in your situation) should be graphed accordingly using the x-axis and the y's (which are 4 and 6 in your situation) should be graph accordingly using the y-axis.

3 0

3 years ago

I dont get this, can anyone help?

balandron [24]

Answer:

this is an x y coordinate plane, start with x and then move to y.

6 0

3 years ago

Teresa drove on monday tuesday and Wednesday. On monday her driving distance was 120 miles. The ratio of Wednesdays distance is

SSSSS [86.1K]

Answer:

Teresa drive <u>90 miles</u> on Tuesday.

Step-by-step explanation:

Given:

Teresa drove on Monday, Tuesday and Wednesday.

On monday her driving distance was 120 miles.

The ratio of Wednesdays distance is 3/5.

The ratio of Wednesdays distance to Tuesday's distance is 5/4.

Now, to find the miles Teresa drive on Tuesday:

Teresa driving distance on Monday was = 120 miles.

Her driving distance on Wednesday is = $\frac{3}{5}\ of\ 120.$

$=\frac{3}{5} \times 120\\\\=0.6\times 120\\\\=72\ miles.$

Now, her driving distance on Tuesday is:

$\frac{5}{4} \ of\ 72\\\\=\frac{5}{4} \times 72\\\\=1.25\times 72\\\\=90\ miles.$

Therefore, Teresa drive 90 miles on Tuesday.

7 0

3 years ago

Square of 5x+1 by 5x

gayaneshka [121]

Answer:

5x(5x+1)^2

= 5x(25x^2+10x+1)

= 125x^3+50x^2+5x

6 0

3 years ago

If a factory continuously dumps pollutants into a river at the rate of the quotient of the square root of t and 45 tons per day,

julsineya [31]

<h2>Hello!</h2>

The answer is:

The first option, the amount dumped after 5 days is 0.166 tons.

To solve the problem, we need to integrate the given expression and evaluate using the given time.

So, integrating we have:

$\int\limits^5_0 {\frac{\sqrt{t} }{45} } \, dt=\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \, dt\\\\\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \ dt=\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt\\\\\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt=(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)\\\\(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)=(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)$

$(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)=(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)\\\\(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)=(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)\\\\(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)=(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})\\\\(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})=\frac{2}{135}*11.18-0=0.1656=0.166$

Hence, we have that the amount dumped after 5 days is 0.166 tons.

Have a nice day!

5 0

3 years ago