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2 years ago

How many pints equal 3 gallons?

Mathematics

1 answer:

qaws [65]2 years ago

3 0

There are 8 pints in 1 gallon. Therefore, 24 pints are equal to 3 gallons.
Hope this helps you! :)

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I really need help with this ... I hope someone can help!

viva [34]

9514 1404 393

Answer:

∠B = 53°

∠C = 127°

∠D = 127°

Step-by-step explanation:

The trapezoid is isosceles, as indicated by the hash marks on sides AD and BC. This means it is symmetrical about the vertical center-line. Angles on the right will have the same measures as the symmetrical angles on the left.

So, you know immediately that ∠B = ∠A = 53°.

Because AB ║ DC, ∠A and ∠D are supplementary. That is, ...

∠D = ∠C = 180° -53°

∠D = ∠C = 127°

4 0

2 years ago

Preston's vegetable patch is 7 yards wide and 17 yards long. Preston wants to build a fence around the vegetable patch. The fenc

bezimeni [28]

Answer: 238$

Step-by-step explanation: 7 x 17 = 119 mutiply this x 2.00 = 238

6 0

2 years ago

CALC- limits please show your method

gladu [14]

A. Factor the numerator as a difference of squares:

$\displaystyle\lim_{x\to9}\frac{x-9}{\sqrt x-3}=\lim_{x\to9}\frac{(\sqrt x-3)(\sqrt x+3)}{\sqrt x-3}=\lim_{x\to9}(\sqrt x+3)=6$

c. As $x\to\infty$ , the contribution of the terms of degree less than 2 becomes negligible, which means we can write

$\displaystyle\lim_{x\to\infty}\frac{4x^2-4x-8}{x^2-9}=\lim_{x\to\infty}\frac{4x^2}{x^2}=\lim_{x\to\infty}4=4$

e. Let's first rewrite the root terms with rational exponents:

$\displaystyle\lim_{x\to1}\frac{\sqrt[3]x-x}{\sqrt x-x}=\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}$

Next we rationalize the numerator and denominator. We do so by recalling

$(a-b)(a+b)=a^2-b^2$
$(a-b)(a^2+ab+b^2)=a^3-b^3$

In particular,

$(x^{1/3}-x)(x^{2/3}+x^{4/3}+x^2)=x-x^3$
$(x^{1/2}-x)(x^{1/2}+x)=x-x^2$

so we have

$\displaystyle\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}\cdot\frac{x^{2/3}+x^{4/3}+x^2}{x^{2/3}+x^{4/3}+x^2}\cdot\frac{x^{1/2}+x}{x^{1/2}+x}=\lim_{x\to1}\frac{x-x^3}{x-x^2}\cdot\frac{x^{1/2}+x}{x^{2/3}+x^{4/3}+x^2}$

For $x\neq0$ and $x\neq1$ , we can simplify the first term:

$\dfrac{x-x^3}{x-x^2}=\dfrac{x(1-x^2)}{x(1-x)}=\dfrac{x(1-x)(1+x)}{x(1-x)}=1+x$

So our limit becomes

$\displaystyle\lim_{x\to1}\frac{(1+x)(x^{1/2}+x)}{x^{2/3}+x^{4/3}+x^2}=\frac{(1+1)(1+1)}{1+1+1}=\frac43$

3 0

2 years ago

13-yard gain please help

fgiga [73]

Answer:

+ 13?

i tried i apologize if its wrong idk what you ment

6 0

3 years ago

Point a pq and r are collinear on pr and pq: pr =2/3. P is located on the origin q is located at (x,y) and r is located at (-12,

dybincka [34]

The values of x and y are 8 and 0 respectively.

Explanation

As 'p' is located at (0,0) and 'r' is located at (-12,0) , that means both 'p' and 'r' are on the x-axis. So point 'q' will be also on the x-axis , and 'q' is located at (x,y)

So, the value of y will be 0. That means the co ordinate of point 'q' is (x, 0)

Now, using the Distance formula, length of 'pq' $= \sqrt{(x-0)^2 +(0-0)^2}= \sqrt{x^2}= x$

and length of 'pr' $= \sqrt{(-12-0)^2+(0-0)^2}= \sqrt{144}=12$

Given that, pq : pr =2/3

So....

$\frac{x}{12} =\frac{2}{3}\\ \\ 3x=24\\ \\ x= \frac{24}{3}=8$

Thus, the values of x and y are 8 and 0 respectively.

8 0

3 years ago