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3 years ago

Please help with this question

Mathematics

1 answer:

VLD [36.1K]3 years ago

3 0

Have you ever heard of slader.com?

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Analyze the given diagram of the carbon cycle below. An image of carbon cycle is shown. The sun, a cloud, two trees, one towards

lianna [129]

Answer:

what the butt is this

Step-by-step explanation:

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3 years ago

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Write the quadratic equation whose roots are 1 and -2, and whose leading coefficient is 3.

kirill115 [55]

Answer:

See below

Step-by-step explanation:

The quadratic equation in factored form:

a(x - α)(x - β), where a- leading coefficient, α and β - roots

We have:

a = 3, α = 1, β = -2

Substitute them to get:

3(x - 1)(x + 2)

Converting this into standard form:

3(x - 1)(x + 2) =
3(x² + x - 2) =
3x² + 3x - 6

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3 years ago

Can someone help me find the surface area of the cube???

nirvana33 [79]

The equation needed is (Surface area of a cube)
A=6×a²
Therefore your answer would be...
194.94 in.³

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3 years ago

Ten points to the correct answer

ale4655 [162]

This should be the correct answer.

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4 years ago

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On the 1st January 2014 Carol invested some money in a bank account.

Ghella [55]

Answer:

$\large \boxed{\text{\pounds 23 360.00}}$

Step-by-step explanation:

The formula for the accrued amount from compound interest is

$A = P \left(1 + \dfrac{r}{n}\right)^{nt}$

1. Amount in account on 1 Jan 2015

(a) Data:

a = £23 517.60

r = 2.5 %

n = 1

t = 1 yr

(b) Calculations:

r = 0.025

$\begin{array}{rcl}23517.60 & = & P\left (1 + \dfrac{r}{n}\right)^{nt}\\\\& = & P\left (1 + \dfrac{0.025}{1}\right)^{1\times1}\\\\& = & P (1 + 0.025)\\ & = & 1.025 P\\P & = & \dfrac{23517.60 }{1.025} \\\\& = & 22 944.00 \\\end{array}$

The amount that gathered interest was £22 944.00 but, before the interest started accruing, Carol had withdrawn £1000 from the account.

She must have had £23 944 in her account on 1 Jan 2015.

(2) Amount originally invested

(a) Data

A = £23 944.00

$\begin{array}{rcl}23 944.00 & = & 1.025 P\\P & = & \dfrac{23 944.00 }{1.025} \\\\& = & \mathbf{23 360.00} \\\end{array}\\\text{Carol originally invested $\large \boxed{\textbf{\pounds23 360.00}}$ in her account.}$

3. Summary

1 Jan 2014 P = £23 360.00

1 Jan 2015 A = 23 944.00

Withdrawal = -1 000.00

P = 22 944.00

1 Jan 2016 A = £23 517.60

5 0

3 years ago