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Oksanka [162]
3 years ago
13

Pls help and have a great day <333!!!!

Mathematics
1 answer:
nirvana33 [79]3 years ago
8 0

Answer:

find the least common multiple for the denominator and multiply with the numerator

for the first one,

Step-by-step explanation:

the common multiple for 4 and 6 is 12

4×3=12

6×2=12

multiply the same numbers with the numerator

1×3=3

5×2=10

the answer is 3/12 and 10/12

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X+y= 2
bija089 [108]

Answer:

11

Step-by-step explanation:

11

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3 years ago
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A right triangle has one leg that measures 7 inches, and the second leg measures 10 inches. How long is the hypotenuse?
faltersainse [42]

Answer:

Hypotenuse≈12.21

Step-by-step explanation:

\sqrt{7^{2} +10^{2} }   ≈12.21

to calculate this you would need to find the square root of the first leg squared added to the second leg squared.

the hypotenuse is the longest side of the triangle

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3 years ago
A bale of hay was 100 cm long by 40 cm wide by 20 cm tall. What is the volume of the bale of hay?
Ksenya-84 [330]

Answer:

80000 cm cubed

Step-by-step explanation:

L*W*H

100*20*20

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2 years ago
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Heyy! please help posted picture of question
Bond [772]
A and D are true



good luck
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3 years ago
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Please solve i will give brainiest 100 point question ****** do the whole page please need to pass or i will fail its my final t
dmitriy555 [2]

Answer:

1. Find the difference between the areas.

<u>Area of the small rectangle</u>: (x+2)(x+7)=x^2+7x+2x+14=x^2+9x+14

<u>Area of the big rectangle</u>: (x+9)(x+11)=x^2+11x+9x+99=x^2+20x+99

The difference is: 11x+85

( x^2+20x+99)- (x^2+9x+14)=x^2+20x+99-x^2-9x-14=11x+85

2.

You can solve this question just by looking at the graph.

a) The height is 4 meters.

f(d)=h=-2d^2+7d+4

To find the height of the bleachers, we should consider the moment before the shoot, when the distance is equal to 0.

f(0)=h=-2(0)^2+7(0)+4

h=4

The height is 4 meters.

b) 9 meters.

For d=1

f(1)=h=-2(1)^2+7(1)+4

f(1)=h=-2+7+4

h=9

b) The ball travels 4 meters.

But to calculate it, it is when h=0

0=-2d^2+7d+4

Using the quadratic formula:

$d=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$d=\frac{-7 \pm \sqrt{7^2-4\left(-2\right)4}}{2\left(-2\right)}$

$d=\frac{-7\pm\sqrt{81}}{-4}$

$d=\frac{-7\pm9}{-4}$

It will give us to solutions, once it is a quadratic equation, but we are talking about a positive distance.

$d=-\frac{1}{2} \text{ or }d=4$

3.

In this question, we have to find the area of the cylinder and the sphere.

From the information given, we have

a = 5mm and d = 5mm, therefore the radius is 2.5 mm.

The volume of a cylinder:

V=\pi r^2h

V=\pi (2.5)^2 \cdot 5

V=31.25 \pi

V_{c} \approx 98.17 \text{ m}^3

The volume of the sphere:

$V=\frac{4}{3}  \pi r^2$

V_{s} \approx 65.4 \text{ m}^3

The volume of the capsule is approximately 163.57  \text{ m}^3

3 0
3 years ago
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