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2 years ago

Please help me with word problems

Mathematics

1 answer:

barxatty [35]2 years ago

7 0

Answer:

Step-by-step explanation:

.16x + 22.92 = 25.00

.16x = 2.08

x = 13 yards of ribbon

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I need help plz i need help

Marrrta [24]

To estimate, you can round each decimal to the nearest whole number, and then add them up together to get a number close to the real answer.

I hope this helps! :)

7 0

3 years ago

Read 2 more answers

Help meeeee!!!!!!!!!!! Also I have more of these sooo

Daniel [21]

x would be 6 because you double 5 to get ten, so you double 3 to get 6

8 0

3 years ago

A farmer has a soybean field with an area of 55 square meters and a cornfield with an area of 58 square meters. he also has a wh

dexar [7]

A farmer has order the fields from least area to greatest area is,

Soybean field < Cornfield < Wheat field.

Based on the given conditions,

A farmer has a soybean field with an area = 55 square meters

A farmer has a corn field with an area = 58 square meters

He also has a wheat field that is 52 times the size of the soybean field.

So,

We can write,

52 times the size of the soybean field.

That means,

= 55 * 52

Calculate the product or quotient,

= 2860

Wheat field = 2860

Hence,

Soybean field = 55

Cornfield = 58

Wheat field = 2860

Soybean field < Cornfield < Wheat field

Therefore,

A farmer has order the fields from least area to greatest area is,

Soybean field < Cornfield < Wheat field.

To learn more about information visit Area problems :

brainly.com/question/16735292

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5 0

1 year ago

Consider xưy" – 14xy' + 56y = 0. Find all values of r such that y=x" satisfies the differential equation for x > 0. Enter as

beks73 [17]

I bet the ODE is supposed to read

$x^2y''-14xy'+56y=0$

Then if $y=x^r$ , we have $y'=rx^{r-1}$ and $y''=r(r-1)x^{r-2}$ , and substituting these into the ODE gives

$r(r-1)x^r-14rx^r+56x^r=0\implies r(r-1)-14r+56=r^2-15r+56=0$

Solving for <em>r</em>, we find

$r^2-15r+56=(r-8)(r-7)=0\implies \boxed{r=8\text{ or }r=7}$

so that $y_1=x^8$ and $y_2=x^7$ are two fundamental solutions to the ODE. Thus the general solution is

$\boxed{y(x)=C_1x^8+C_2x^7}$

Given that $y(1)=4$ and $y'(1)=3$ , we get

$\begin{cases}4=C_1+C_2\\3=8C_1+7C_2\end{cases}\implies C_1=-25\text{ and }C_2=29$

So the particular solution is

$\boxed{y(x)=29x^7-25x^8}$

4 0

4 years ago

FLVS STUDENTS !!! For the 5.05 Two-Variable Linear Inequalities algebra 1 assignment can someone send me there answers ?!?!?!

iogann1982 [59]

Degree is the value of the highest power of the variable, and hence in this case is 8

6 0

4 years ago