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irinina [24]
3 years ago
7

ANSWER ASAP!!! PLEASE... I really need an answer ;-; ..

Mathematics
2 answers:
Kisachek [45]3 years ago
4 0
I know I got my answer deleted the first time but I'm answering again cause it's worth 25 points
arsen [322]3 years ago
3 0

1:59 pm is 13:59, 13:59 - 06:31 = 07:28

Answer: 7 hours 28 minutes to reply

2.,3. On a "date" at 6:31 am when he already has a girlfriend?   Dump him and don't look back.



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Determine if the statement is always, sometimes, or never true. Two angles that are vertical are __________ adjacent.
Morgarella [4.7K]

Never true. Vertical angles are equal to each other, not adjacent to each other.

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4 years ago
a rectangular cake has a length of 15 inches and a width of 10 inches how many whole square pieceslenghts of 2 1/2 inches can be
sineoko [7]

Answer:

24

Step-by-step explanation:

The cake measures 10 inches by 15 inches. If pieces are 2.5 inches long and wide (they are square and equal) then 6 pieces will be cut along the length.

15/ 2.5 = 6

And 4 pieces will be cut along the width.

10/ 2.5 = 4

All together when the entire cake is cut, there will be 4*6 = 24 squares 2.5 inch long pieces.

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4 years ago
What ordered pair is a solution to this equation 4x+y=20
Ahat [919]

Answer:

happy holidays!!!!!!!! love you u are strong and a true buddy

Step-by-step explanation:

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3 years ago
Which of the following is true about a linear function? (4 points)
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please urgent please beguvtıvıgvıgıgugyyy ve teşekküre basarsan gider gitmez bu soruya
7 0
3 years ago
99 POINT QUESTION, PLUS BRAINLIEST!!!
Elden [556K]
We draw region ABC. Lines that connect y = 0 and y = x³ are vertical so:
(i) prependicular to the axis x - disc method;
(ii) parallel to the axis y - shell method;
(iii) parallel to the line x = 18 - shell method.

Limits of integration for x are easy x₁ = 0 and x₂ = 9.
Now, we have all information, so we could calculate volume.

(i)

V=\pi\cdot\int\limits_a^bf^2(x)\, dx\qquad\implies \qquad a=0\qquad b=9\qquad f(x)=x^3


V=\pi\cdot\int\limits_0^9(x^3)^2\, dx=\pi\cdot\int\limits_0^9x^6\, dx=\pi\cdot\left[\dfrac{x^7}{7}\right]_0^9=\pi\cdot\left(\dfrac{9^7}{7}-\dfrac{0^7}{7}\right)=\dfrac{9^7}{7}\pi=\\\\\\=\boxed{\dfrac{4782969}{7}\pi}

Answer B. or D.

(ii)

V=2\pi\cdot\int\limits_a^bx\cdot f(x)\, dx


V=2\pi\cdot\int\limits_0^{9}(x\cdot x^3)\, dx=2\pi\cdot\int\limits_0^{9}x^4\, dx=
2\pi\cdot\left[\dfrac{x^5}{5}\right]_0^9=2\pi\cdot\left(\dfrac{9^5}{5}-\dfrac{0^5}{5}\right)=\\\\\\=2\pi\cdot\dfrac{9^5}{5}=\boxed{\dfrac{118098}{5}\pi}

So we know that the correct answer is D.

(iii)
Line x = h

V=2\pi\cdot\int\limits_a^b(h-x)\cdot f(x)\, dx\qquad\implies\qquad h=18


V=2\pi\cdot\int\limits_0^9\big((18-x)\cdot x^3\big)\, dx=2\pi\cdot\int\limits_0^9(18x^3-x^4)\, dx=\\\\\\=2\pi\cdot\left(\int\limits_0^918x^3\, dx-\int\limits_0^9x^4\, dx\right)=2\pi\cdot\left(18\int\limits_0^9x^3\, dx-\int\limits_0^9x^4\, dx\right)=\\\\\\=2\pi\cdot\left(18\left[\dfrac{x^4}{4}\right]_0^9-\left[\dfrac{x^5}{5}\right]_0^9\right)=2\pi\cdot\Biggl(18\biggl(\dfrac{9^4}{4}-\dfrac{0^4}{4}\biggr)-\biggl(\dfrac{9^5}{5}-\dfrac{0^5}{5}\biggr)\Biggr)=\\\\\\

=2\pi\cdot\left(18\cdot\dfrac{9^4}{4}-\dfrac{9^5}{5}\right)=2\pi\cdot\dfrac{177147}{10}=\boxed{\dfrac{177147\pi}{5}}

Answer D. just as before.

6 0
3 years ago
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