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2 years ago

PLEASE ANSWER ASAP!!! will mark brainliest

Mathematics

1 answer:

rewona [7]2 years ago

5 0

Answer:

$a^{4} b^{6}$

Step-by-step explanation:

ab² * ab³ * a²b

a^4 b^6

$a^{4} b^{6}$

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What is a polygonomial

Nadya [2.5K]

Answer:

See below

Step-by-step explanation:

Polynomials are sums of terms in the form of k⋅xⁿ, where k is any number and n is a positive integer. For example, 3x²+2x-5 is a polynomial.

3 0

2 years ago

On a certain hot summer's day, 581 people used the public swimming pool. The daily prices are $1.50 for children and $2.50 for

Tanya [424]

Answer:

A = 219

C = 362

Step-by-step explanation:

Given,

581 = C + A

and

$1090.50 = 1.5C + 2.5A

solve 1st for A

A =581 - C

substute into 2nd

1090.50 = 1.5C + 2.5(581 - C)

solve for C

1090.50 = 1.5C + 1452.50 - 2.5C

1090.50 = -C + 1452.50

1090.50 - 1452.50 = -C

-362 = -C

C = 362

then

581 = C + A

581 = 362 + A

219 = A

A = 219

8 0

3 years ago

What single decimal multiplier would you use to increase by 10% followed by a 16% decrease?

Neko [114]

So a decrease of 16% followed by an increase of 14% is the same as multiplying by 0.9576 (which is a decrease of 4.24%)

7 0

3 years ago

Evaluate the expression below if x = 3 and y= -4.<br> X - Y - 3+ (-57)

Firdavs [7]

Answer:

-53

Step-by-step explanation:

If x=3 and y=-4

Plug the numbers into the equation

3 - (-4) - 3 + (-57)

Now solve the equation right to left

3 - (-4) = 7

7 - 3 = 4

4 - 57 = -53

So the answer is -53

4 0

3 years ago

Read 2 more answers

The position of an object moving along an x axis is given by x = 3.24 t - 4.20 t2 + 1.07 t3, where x is in meters and t in secon

Pavel [41]

Answer and explanation:

Given : The position of an object moving along an x axis is given by $x=3.24t-4.20t^2+1.07t^3$ where x is in meters and t in seconds.

To find : The position of the object at the following values of t :

a) At t= 1 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(1)=3.24(1)-4.20(1)^2+1.07(1)^3$

$x(1)=3.24-4.20+1.07$

$x(1)=0.11$

b) At t= 2 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(2)=3.24(2)-4.20(2)^2+1.07(2)^3$

$x(2)=6.48-16.8+8.56$

$x(2)=-1.76$

c) At t= 3 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(3)=3.24(3)-4.20(3)^2+1.07(3)^3$

$x(3)=9.72-37.8+28.89$

$x(3)=0.81$

d) At t= 4 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(4)=3.24(4)-4.20(4)^2+1.07(4)^3$

$x(4)=12.96-67.2+68.48$

$x(4)=14.24$

(e) What is the object's displacement between t = 0 and t = 4 s?

At t=0, x(0)=0

At t=4, x(4)=14.24

The displacement is given by,

$\triangle x=x(4)-x(0)$

$\triangle x=14.24-0$

$\triangle x=14.24$

(f) What is its average velocity from t = 2 s to t = 4 s?

At t=2, x(2)=-1.76

At t=4, x(4)=14.24

The average velocity is given by,

$\triangle x=x(4)-x(2)$

$\triangle x=14.24-(-1.76)$

$\triangle x=14.24+1.76$

$\triangle x=16$

4 0

3 years ago