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Kamila [148]
2 years ago
12

Find the total surface area for this triangular prism height - 24cm

Mathematics
1 answer:
Andrej [43]2 years ago
7 0

Answer:

24 cm² is the correct answer

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Mr. Gail uses 12 gallons of gas for a 400-mile trip. How far can I travel on one gallon of gas​
bulgar [2K]

Answer:

100/3 or 33.34 miles a gallon

Step-by-step explanation:

Divide the number of miles by gallons of gas to get the number of miles traveled on one gallon of gas.

3 0
2 years ago
Write the contrapositive of the conditional statement. Determine whether the contrapositive is true or false. If it is false, fi
Vikentia [17]

Answer:

D is the contrapositive.

Step-by-step explanation:

Contrapositive of if A then B is if not B then not A

3 0
3 years ago
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Pattern A follows the rule "subtract 3" and Pattern B follows the rule "subtract 4." Pattern A 20 17 14 11 8 Pattern B 24 20 16
ruslelena [56]

Answer:

(20,24) (17,20) and (8,8)

Hope this helps

Step-by-step explanation:

8 0
3 years ago
Find the inverse of the given function.<br><br> f(x) = -1/2√x + 3, x ≥ -3
inna [77]
<span>Find the inverse of the given function.

f(x) = -1/2√x + 3, x ≥ -3

I will have to assume that you meant f(x) = -(1/2)sqrt(x) + 3.  If you actually meant f(x) = -(1/2)sqrt(x+3), then obviously the correct result would be different.

1.  Replace "f(x)" by "y:"  y </span>= -(1/2)sqrt(x) + 3
 2.  Interchange x and y:  x = -(1/2)sqrt(y) + 3
3.  Solve for y:  x-3=-(1/2)sqrt(y), so that 2(3-x)= sqrt(y) and y=+sqrt(2[3-x])
4.  Replace "y" with 

           -1
        f      (x) = sqrt(2[3-x])

Here, there are restrictions on x, since the domain of the sqrt function does not include - numbers.  The domain here is (-infinity,3]
4 0
3 years ago
Read 2 more answers
Evaluate the integral. W (x2 y2) dx dy dz; W is the pyramid with top vertex at (0, 0, 1) and base vertices at (0, 0, 0), (1, 0,
In-s [12.5K]

Answer:

\mathbf{\iiint_W (x^2+y^2) \ dx \ dy \ dz = \dfrac{2}{15}}

Step-by-step explanation:

Given that:

\iiint_W (x^2+y^2) \ dx \ dy \ dz

where;

the top vertex = (0,0,1) and the  base vertices at (0, 0, 0), (1, 0, 0), (0, 1, 0), and (1, 1, 0)

As such , the region of the bounds of the pyramid is: (0 ≤ x ≤ 1-z, 0 ≤ y ≤ 1-z, 0 ≤ z ≤ 1)

\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \int ^{1-z}_0 \int ^{1-z}_0 (x^2+y^2) \ dx \ dy \  dz

\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \int ^{1-z}_0 ( \dfrac{(1-z)^3}{3}+ (1-z)y^2) dy \ dz

\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0  \ dz \  ( \dfrac{(1-z)^3}{3} \ y + \dfrac {(1-z)y^3)}{3}] ^{1-x}_{0}

\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0  \ dz \  ( \dfrac{(1-z)^4}{3}+ \dfrac{(1-z)^4}{3}) \ dz

\iiint_W (x^2+y^2) \ dx \ dy \ dz =\dfrac{2}{3} \int^1_0 (1-z)^4 \ dz

\iiint_W (x^2+y^2) \ dx \ dy \ dz =- \dfrac{2}{15}(1-z)^5|^1_0

\mathbf{\iiint_W (x^2+y^2) \ dx \ dy \ dz = \dfrac{2}{15}}

7 0
3 years ago
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