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Svet_ta [14]
3 years ago
8

Please help I am in need of desperate help

Mathematics
1 answer:
KengaRu [80]3 years ago
5 0

Answer:

1+2=\frac{9}{3}

Step-by-step explanation:

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Freshman Year
prohojiy [21]
I think it b t up Orlando
7 0
3 years ago
A group of 32 players forms 4 volleyball teams if there are 96 players how many teams can be formed
kotegsom [21]

Answer:

12 teams

Step-by-step explanation:

We know that 38 players can fit into 4 groups, 8 times ...

32 divide 8 = 4

We have the number with 96, and we need to find the groups

we do the same as done with the first one...

because 32 divided by 8 is 4

we do

96 divide 8 = 12

aswell

<h2><u><em>hope this helped : )</em></u></h2>
6 0
3 years ago
Read 2 more answers
Please help with algebra problem
kumpel [21]
Number of weekend minutes used: x
Number of weekday minutes used: y

This month Nick was billed for 643 minutes:
(1) x+y=643

The charge for these minutes was $35.44
Telephone company charges $0.04 per minute for weekend calls (x)
and $0.08 per minute for calls made on weekdays (y)
(2) 0.04x+0.08y=35.44

We have a system of 2 equations and 2 unkowns:
(1) x+y=643
(2) 0.04x+0.08y=35.44

Using the method of substitution
Isolating x from the first equation:
(1) x+y-y=643-y
(3) x=643-y

Replacing x by 643-y in the second equation
(2) 0.04x+0.08y=35.44
0.04(643-y)+0.08y=35.44
25.72-0.04y+0.08y=35.44
0.04y+25.72=35.44

Solving for y:
0.04y+25.72-25.72=35.44-25.72
0.04y=9.72

Dividing both sides of the equation by 0.04:
0.04y/0.04=9.72/0.04
y=243

Replacing y by 243 in the equation (3)
(3) x=643-y
x=643-243
x=400

Answers:
The number of weekends minutes used was 400
The number of weekdays minutes used was 243
6 0
3 years ago
Last year's sales were $13,500. Sales have increased 47% this year. How much is the increase?
IRISSAK [1]
47% of 13,500 is 6,345. So this years sales would be $19,845.
8 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=&#10;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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