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3 years ago

-4 times a number plus 29

Mathematics

1 answer:

kipiarov [429]3 years ago

8 0

-4x+29? I think that's what you're asking.

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Mrs. Cerra planted 4 ⅞ acres of wheat and 1 ⅝ acres of corn. How much more wheat did she plant than corn?

alukav5142 [94]

We can subtract the acres of corn from the acres of wheat.
$4 \frac{7}{8} - 1 \frac{5}{8} = 3 \frac{2}{8} = 3 \frac{1}{4}$
She planted 3 1/4 more acres of wheat than corn.

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3 years ago

What is the slope of line on the graph help pwease

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Answer: is where that one dot is at home n the y-axis

Step-by-step explanation:

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3 years ago

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In 2010, the number of houses built in Town A was 25 percent greater than the number of houses built in Town B. If 70 houses wer

Stella [2.4K]

Answer:

The number of houses built in Town B is 56.

Step-by-step explanation:

We are given that in 2010, the number of houses built in Town A was 25 percent greater than the number of houses built in Town B.

Also, 70 houses were built in Town A during 210.

Let the number of houses built in Town B be 'x'.

So, according to the question;

Number of houses built in Town A = Number of houses built in Town B + 25% of the houses built in Town B

$70 = x + (25\% \times x)$

$70 = x(1+0.25)$

$x=\frac{70}{1.25}$

x = 56

Hence, the number of houses built in Town B is 56.

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3 years ago

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The amount that gathered interest was £22 944.00 but, before the interest started accruing, Carol had withdrawn £1000 from the account.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