All categories

2 years ago

Simplify: (–2x+y)(-58 y) O 10x10, 10xv2 10xy 10x10 ,

Mathematics

1 answer:

alekssr [168]2 years ago

8 0

Answer:

Choice C

Step-by-step explanation:

$(-2x^2y)(-5x^5y^3)=\\\\(-2)(-5)\times (x^2)(x^5)\times (y)(y^3)=\\\\10x^7y^4$

Hope this helps!

You might be interested in

The cost per ounce of a drink, c, varies inversely as the number of ounces, n. Six ounces of the drink costs 60 cents per ounce.

kaheart [24]

Answer:

c = 360/n

Step-by-step explanation:

From the question, we are told that:

The cost per ounce of a drink, c, varies inversely as the number of ounces, n.

Mathematically this is expressed as:

c ∝ 1/n

Using the constant of proportionality k

c = k/n

Six ounces of the drink costs 60 cents per ounce.

c =60 cents per ounce

n = 6 ounces

Mathematically we have

c = k/n

60 = k/6

Cross multiply to find the constant k

k = 60 × 6

k = constant of proportionality = 360

Therefore, using the mathematically expression "c = k/n" the equation that represents this relationship is

c = 360/n

3 0

2 years ago

The cost of 5 cans of dog food is $4.35. At this price, how much does 24 cans of dog food cost?

Harrizon [31]

Answer:

$20.88

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

I urgently need the answer plzzzzzzz with explanation too plzzzzz

Eddi Din [679]

The pizza is fresh and the doe is fresh made

3 0

3 years ago

Read 2 more answers

The function below is written in vertex form or intercept form. Rewrite them in standard form and show your work.

lora16 [44]

Answer:

The standard form as $y=5x^2+30x+41$

Step-by-step explanation:

Given: A function which is written in vertex form or intercept form.

We have to re-write it in standard form that in terms of

Given $y = 5(x+3)^2-4$

Squaring using $(a+b)^2=a^2+b^2+2ab$ , we get,

$y=5(x^2+9+6x)-4$

Multiply 5 inside , we get,

$y=5x^2+45+30x-4$

Solving further , we get,

$y=5x^2+30x+41$

Thus , we have obtained the standard form as $y=5x^2+30x+41$

8 0

2 years ago

Read 2 more answers

Consider a series circuit consisting of a resistor of R ohms, an inductor of L henries, and variable voltage source of V(t) volt

ser-zykov [4K]

Answer:

$I(t)=\frac{1}{3}(1-e^{-30t})$

Step-by-step explanation:

We are given that

$\frac{dI}{dt}+\frac{R}{L}I=\frac{V(t)}{L}$

R=150 ohm

L=5 H

V(t)=10 V

$P=\frac{R}{L}=\frac{150}{5}=30$

$I.F=e^{\int Pdt}=e^{\int 30 dt}=e^{30 t}$

$I(t)\times I.F=\int e^{30 t}\times 10 dt+C$

$I(t)\times e^{30 t}=\frac{10}{30}e^{30 t}+C$

$I(t)=\frac{1}{3}+Ce^{-30 t}$

I(0)=0

Substitute t=0

$0=\frac{1}{3}+C$

$C=-\frac{1}{3}$

Substitute the values

$I(t)=\frac{1}{3}-\frac{1}{3}e^{-30 t}$

$I(t)=\frac{1}{3}(1-e^{-30t})$

5 0

3 years ago