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3 years ago

How many cups are in a 13 gallons

Mathematics

2 answers:

djyliett [7]3 years ago

7 0

<em>there are 208 cups in 13 gallons.</em>

laila [671]3 years ago

3 0

✧･ﾟ: *✧･ﾟ:* *:･ﾟ✧*:･ﾟ✧

Hello!

✧･ﾟ: *✧･ﾟ:* *:･ﾟ✧*:･ﾟ✧

❖ 208 cups

1 gal = 16 cups

16 x 13 = 208 cups

~ ʜᴏᴘᴇ ᴛʜɪꜱ ʜᴇʟᴘꜱ! :) ♡

~ ᴄʟᴏᴜᴛᴀɴꜱᴡᴇʀꜱ

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The measures of three angles of a triangle are given by (8x - 5), (2x)º, and (3x – 10). What is the measure of the largest angle

zhannawk [14.2K]

Answer: 115°

Step-by-step explanation:

We should note that the sum of the angles that are in a triangle is equal to 180°, therefore we will add the angles given and equate them to 180°. This will be:

(8x - 5) + (2x) + (3x - 10) = 180

8x - 5 + 2x + 3x - 10 = 180

Collect like terms

8x + 2x + 3x = 180 + 5 + 10

13x = 195

x = 195/13

x = 15

The angles are:

8x - 5 = 8(15) - 5 = 120 - 5 = 115°

2x = 2(15) = 30°

3x - 10 = 3(15) - 10 = 45 - 10 = 35°

Therefore, the largest angle is 115°

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3 years ago

What is the length of a leg of a right triangle if the hypotenuse is 9 inches and the other leg is 8 inches.

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Answer:

√17 or 4.12 inches

Step-by-step explanation:

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or 4.12 inches

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3 years ago

Find the slope from the table:

Aleonysh [2.5K]

Answer:

B! $\frac{-5}{3}$

Step-by-step explanation:

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2 years ago

In Exercises 40-43, for what value(s) of k, if any, will the systems have (a) no solution, (b) a unique solution, and (c) infini

svet-max [94.6K]

Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

The system cannot have unique solution.

Step-by-step explanation:

We have the following system of equations

$x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2$

The augmented matrix is

$\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]$

The reduction of this matrix to row-echelon form is outlined below.

$R_2\rightarrow R_2-R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]$

$R_3\rightarrow R_3-2R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]$

$R_3\rightarrow R_3-R_2$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]$

The last row determines, if there are solutions or not. To be consistent, we must have k such that

$k^2-k-2=0$

$\left(k+1\right)\left(k-2\right)=0\\k=-1,\:k=2$

Case k = −1:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]$

If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]$

This gives the infinite many solution.

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3 years ago

If a family spend $840 in a year on rent how much do they spend in a manth

dalvyx [7]

Month*
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as 1 year has 12 months
and you get 70$

so the family spends $70 every month

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3 years ago