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aleksley [76]
3 years ago
10

If Jay had 5 cards and Jill had 10 cards. What is the product?

Mathematics
1 answer:
cupoosta [38]3 years ago
5 0
Hello there,

The word "product" mean's to multiply.

So we are going to do 5 x 10.

This will give us 50 because when we do
10+10+10+10+10 is give's us 50. That is 
what multiplication mean's.

Hope this helps

Your correct answer will be 50 card's

~Jurgen
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In ΔEFG, the measure of ∠G=90°, EG = 11, FE = 61, and GF = 60. What ratio represents the sine of ∠F?
lapo4ka [179]

Answer:

sinF = \frac{11}{61}

Step-by-step explanation:

sinF = \frac{opposite}{hypotenuse} = \frac{EG}{FE} = \frac{11}{61}

6 0
3 years ago
Point O is the center of the circle. What is the value of x?
Bad White [126]
OQ = 8 + 9 = 17
x^2 = 17^2 - 8^2
x^2 = 289 - 64
x^2 = 225
x = 15
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HELP ASAP, LINKS AND ABSURD ANSWERS WILL BE REPORTED! I WILL MARK BRAINLIEST!!
Assoli18 [71]

Answer:

1/10th?

Pretty sure that's it.

Correct me if I'm wrong,

4 0
3 years ago
A seal dove 95 meters below the surface of the water to reach a fish. It traveled 18% of the total distance every second.
BlackZzzverrR [31]
95*18%
18%=0.18
95 x 0.18 =17.1
Now, multiply 17.1 by 5 = 85.5
4 0
3 years ago
Jorge is asked to build a box in the shape of a rectangular prism. The maximum girth of the box is 20 cm. What is the width of t
MariettaO [177]

Answer:

The width of the box is 6.7 cm

The maximum volume is 148.1 cm³

Step-by-step explanation:

The given parameters of the box Jorge is asked to build are;

The maximum girth of the box = 20 cm

The nature of the sides of the box = 2 square sides and 4 rectangular sides

The side length of square side of the box = w

The length of the rectangular side of the box = l

Therefore, we have;

The girth = 2·w + 2·l = 20 cm

∴ w + l = 20/2 = 10

w + l = 10

l = 10 - w

The volume of the box, V = Area of square side × Length of rectangular side

∴ V = w × w × l = w × w × (10 - w)

V = 10·w² - w³

At the maximum volume, we have;

dV/dw = d(10·w² - w³)/dw = 0

∴ d(10·w² - w³)/dw = 2×10·w - 3·w² = 0

2×10·w - 3·w² = 20·w - 3·w² = 0

20·w - 3·w² = 0 at the maximum volume

w·(20 - 3·w) = 0

∴ w = 0 or w = 20/3 = 6.\overline 6

Given that 6.\overline 6 > 0, we have;

At the maximum volume, the width of the block, w = 6.\overline 6 cm ≈ 6.7 cm

The maximum volume, V_{max}, is therefore given when w = 6.\overline 6 cm = 20/3 cm  as follows;

V = 10·w² - w³

V_{max} = 10·(20/3)² - (20/3)³ = 4000/27 = 148.\overline {148}

The maximum volume, V_{max} = 148.\overline {148} cm³ ≈ 148.1 cm³

Using a graphing calculator, also, we have by finding the extremum of the function V = 10·w² - w³, the coordinate of the maximum point is (20/3, 4000/27)

The width of the box is;

6.7 cm

The maximum volume is;

148.1 cm³

5 0
3 years ago
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