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andrew11 [14]
3 years ago
14

Find the number of 1 cm cubes that can be placed in the box that right

Mathematics
2 answers:
Arte-miy333 [17]3 years ago
6 0

Answer:

Length x Width x Height = volume

5 x 5 = 25 x 3 = 75

Step-by-step explanation:

Hope this helps can I have brainliest

Inessa [10]3 years ago
3 0

Answer:

75

Step-by-step explanation:

volume of box = LWH = 5 cm * 5 cm * 3 cm = 75 cm^3

volume of a 1-cm cube = LWH = 1 cm * 1 cm * 1 cm = 1 cm^3

number of 1-cm cubes that fit in box = (volume of box)/(volume of 1-cm cube)

number = (75 cm^3)/(1 cm^3) = 75

Answer: 75

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Pls help will mark as brainliest
Vesna [10]

Answer:

C. \frac{60}{100}=\frac{x}{25}

Step-by-step explanation:

Since 40% of the students prefer chocolate ice cream, 60% don't prefer chocolate ice cream. Use this proportion set equal to 25 to find the number of students who prefer vanilla ice cream, instead of chocolate ice cream.

6 0
3 years ago
Read 2 more answers
Evaluate the double integral.
Fynjy0 [20]

Answer:

\iint_D 8y^2 \ dA = \dfrac{88}{3}

Step-by-step explanation:

The equation of the line through the point (x_o,y_o) & (x_1,y_1) can be represented by:

y-y_o = m(x - x_o)

Making m the subject;

m = \dfrac{y_1 - y_0}{x_1-x_0}

∴

we need to carry out the equation of the line through (0,1) and (1,2)

i.e

y - 1 = m(x - 0)

y - 1 = mx

where;

m= \dfrac{2-1}{1-0}

m = 1

Thus;

y - 1 = (1)x

y - 1 = x ---- (1)

The equation of the line through (1,2) & (4,1) is:

y -2 = m (x - 1)

where;

m = \dfrac{1-2}{4-1}

m = \dfrac{-1}{3}

∴

y-2 = -\dfrac{1}{3}(x-1)

-3(y-2) = x - 1

-3y + 6 = x - 1

x = -3y + 7

Thus: for equation of two lines

x = y - 1

x = -3y + 7

i.e.

y - 1 = -3y + 7

y + 3y = 1 + 7

4y = 8

y = 2

Now, y ranges from 1 → 2 & x ranges from y - 1 to -3y + 7

∴

\iint_D 8y^2 \ dA = \int^2_1 \int ^{-3y+7}_{y-1} \ 8y^2 \ dxdy

\iint_D 8y^2 \ dA =8 \int^2_1 \int ^{-3y+7}_{y-1} \ y^2 \ dxdy

\iint_D 8y^2 \ dA =8 \int^2_1  \bigg ( \int^{-3y+7}_{y-1} \ dx \bigg)   dy

\iint_D 8y^2 \ dA =8 \int^2_1  \bigg ( [xy^2]^{-3y+7}_{y-1} \bigg ) \ dy

\iint_D 8y^2 \ dA =8 \int^2_1  \bigg ( [y^2(-3y+7-y+1)]\bigg ) \ dy

\iint_D 8y^2 \ dA =8 \int^2_1  \bigg ([y^2(-4y+8)] \bigg ) \ dy

\iint_D 8y^2 \ dA =8 \int^2_1  \bigg ( -4y^3+8y^2 \bigg ) \ dy

\iint_D 8y^2 \ dA =8 \bigg [\dfrac{ -4y^4}{4}+\dfrac{8y^3}{3} \bigg ]^2_1

\iint_D 8y^2 \ dA =8 \bigg [ -y^4+\dfrac{8y^3}{3} \bigg ]^2_1

\iint_D 8y^2 \ dA =8 \bigg [ -2^4+\dfrac{8(2)^3}{3} + 1^4- \dfrac{8\times (1)^3}{3}\bigg]

\iint_D 8y^2 \ dA =8 \bigg [ -16+\dfrac{64}{3} + 1- \dfrac{8}{3}\bigg]

\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{64-8}{3}\bigg]

\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{56}{3}\bigg]

\iint_D 8y^2 \ dA =8 \bigg [  \dfrac{-45+56}{3}\bigg]

\iint_D 8y^2 \ dA =8 \bigg [  \dfrac{11}{3}\bigg]

\iint_D 8y^2 \ dA = \dfrac{88}{3}

4 0
3 years ago
What is the value of x in 15x-10= 20
Brut [27]
15x-10=20
+10 +10
15x=30
\15 /15
x=2

you add 10 to both -10 (they cancek eachother out) and 20 (equals 30)
Then you divide 30 by 15 (equals 2) and 15 by 15 (they cancel eachother out)
Then you get x=2
4 0
3 years ago
Read 2 more answers
4x^2-5x-15=0 how to find the vertex and y intercept
mezya [45]
ax^2+bx+c=0\\\\the\ vertex:\left(\frac{-b}{2a};\frac{-(b^2-4ac)}{4a}\right)\\\\y-intrcept=c\\=======================================\\\\4x^2-5x-15=0\\\\a=4;\ b=-5;\ c=-15\\\\\frac{-b}{2a}=\frac{-(-5)}{2\cdot4}=\frac{5}{8};\ \frac{-(b^2-4ac)}{4a}=\frac{-[(-5)^2-4\cdot4\cdot(-15)]}{4\cdot4}=-\frac{265}{16}\\\\\boxed{the\ vertex:\left(\frac{5}{8};-\frac{265}{16}\right)}\\\\\boxed{y-intercept:-15}
5 0
3 years ago
A survey of the whole United States shows that 75% of 6th grade students like sports that involve running, that 60% of boys like
slamgirl [31]

Answer: 75% of 6th Grade Students

Step-by-step explanation:

This sample makes the most sense because it is a majority of everyone, not just a majority of certain people. Also, Mountain School would get fewer complaints from other people if they went with a majority of 75%. Finally, running sports usually cost less then other sports.

3 0
2 years ago
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