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3 years ago

. Graham bought 45 bags of potatoes. Each bag weighed exactly 4pounds.How many pounds of potatoes did he buy?

Mathematics

1 answer:

Serggg [28]3 years ago

8 0

Multiply 45 by 4
=180 pounds of potatoes

You might be interested in

Rewrite the equation in ax+by=c form. <br> Y+3=2(x-5)

Deffense [45]

Y+3=2(x-5)

y+3=2x-10

y+3+10=2x

y+13=2x

2x-y=13

6 0

3 years ago

Nth term of this sequence 1, 10, 19, 28

mestny [16]

Answer:

The nth term of the sequence is

Step-by-step explanation:

The sequence above is an arithmetic sequence

For an nth term in an arithmetic sequence

A(n) = a + ( n - 1)d

where a is the first term

n is the number of terms

d is the common difference

From the question

a = 1

d = 10 - 1 = 9 or 19 - 10 = 9 or 28 - 19 = 9

So the nth term of the sequence is

A(n) = 1 + (n - 1)9

= 1 + 9n - 9

= 9n - 8

Hope this helps you

3 0

3 years ago

If a line is 10 feet long but needs to be reduced to 80% of its original, how long will the new line be?

motikmotik

Answer:

8 feet

Step-by-step explanation:

6 0

3 years ago

Which value is NOT a solution of 8x3 – 1 = 0?

Tpy6a [65]

<u><em>Note: As you may have unintentionally missed to add the value choices. But, I would make sure to explain the concept so that you may improve your understanding in terms of solving these type of questions.</em></u>

Answer:

Any value other than the values $x=\frac{1}{2},\:x=-\frac{1}{4}+i\frac{\sqrt{3}}{4},\:x=-\frac{1}{4}-i\frac{\sqrt{3}}{4}$ will not be a solution of $8x^3\:-\:1\:=\:0$ .

Step-by-step explanation:

Considering the equation

$8x^3\:-\:1\:=\:0$

Steps to solve the equation

$8x^3-1=0$

$\mathrm{Add\:}1\mathrm{\:to\:both\:sides}$

$8x^3-1+1=0+1$

$\mathrm{Simplify}$

$x^3=\frac{1}{8}$

$\mathrm{Divide\:both\:sides\:by\:}8$

$\frac{8x^3}{8}=\frac{1}{8}$

$\mathrm{Simplify}$

$x^3=\frac{1}{8}$

$\mathrm{For\:}x^3=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt[3]{f\left(a\right)},\:\sqrt[3]{f\left(a\right)}\frac{-1-\sqrt{3}i}{2},\:\sqrt[3]{f\left(a\right)}\frac{-1+\sqrt{3}i}{2}$

$x=\sqrt[3]{\frac{1}{8}},\:x=\sqrt[3]{\frac{1}{8}}\frac{-1+\sqrt{3}i}{2},\:x=\sqrt[3]{\frac{1}{8}}\frac{-1-\sqrt{3}i}{2}$

So,

$x=\frac{1}{2},\:x=-\frac{1}{4}+i\frac{\sqrt{3}}{4},\:x=-\frac{1}{4}-i\frac{\sqrt{3}}{4}$

Therefore,

Any value other than the values $x=\frac{1}{2},\:x=-\frac{1}{4}+i\frac{\sqrt{3}}{4},\:x=-\frac{1}{4}-i\frac{\sqrt{3}}{4}$ will not be a solution of $8x^3\:-\:1\:=\:0$ .

Keywords: solution, value

Learn more about equation solution from brainly.com/question/1679491

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7 0

3 years ago

Kendra is buying gifts for 5 friends.

Blizzard [7]

Answer:

She could spend $18 on each for the 4 friends

Step-by-step explanation:

100-28= 72

72/4=18

4 0

3 years ago