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3 years ago

40 POINTS!!! PLEASE ANSWER SOON. WILL MARK BRAINLIEST

Mathematics

1 answer:

algol [13]3 years ago

4 0

Answer:

Step-by-step explanation:

N = 6

S = 0 n = 1

2(1) - 1 = 1

S = 0 + 1 = 1

n = n + 1

n = 1 + 1

n = 2

S = 1 n = 2

2(2) - 1 = 3

S = 1 + 3 = 4

n = n + 1

n = 2 + 1

n = 3

S = 4 n = 3

2(3) - 1 = 5

S = 4 + 5 = 9

n = n + 1

n = 3 + 1

n = 4

S = 9 n = 4

2(4) - 1 = 7

S = 9 + 7 = 16

n = n + 1

n = 4 + 1

n = 5

S = 16 n = 5

2(5) - 1 = 9

S = 16 + 9 = 25

n = n + 1

n = 5 + 1

n = 6

S = 25 n = 6

2(6) - 1 = 11

S = 25 + 11 = 36

n = n + 1

n = 6 + 1

n = 7 > N

Stop

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Sanam eats 5 almonds daily. How many almonds does she eat in 157 days?

Ede4ka [16]

Answer:

785 almonds

Step-by-step explanation:

157 x 5 = 785

8 0

2 years ago

The diameter of a tire is 2.5 ft. Use this measurement to answer parts a and b. Show all work to receive full credit

Alla [95]

Answer:

The answer is below

Step-by-step explanation:

The diameter of a tire is 2.5 ft. a. Find the circumference of the tire. b. About how many times will the tire have to rotate to travel 1 mile?

Solution:

a) The circumference of a circle is the perimeter of the circle. The circumference of the circle is the distance around a circle, that is the arc length of the circle. The circumference of a circle is given by:

Circumference = 2π × radius; but diameter = 2 × radius. Hence:

Circumference = π * diameter.

Given that diameter of the tire = 2.5 ft:

Circumference of the tire = π * diameter = 2.5 * π = 7.85 ft

b) since the circumference of the tire is 7.85 ft, it means that 1 revolution of the tire covers a distance of 7.85 ft.

1 mile = 5280 ft

The number of rotation required to cover 1 mile (5280 ft) is:

number of rotation = $\frac{5280\ ft}{7.85\ ft\ per\ rotation}=672\ rotations\\\\\\$

4 0

3 years ago

Read 2 more answers

Y=-x^2+2x+10 y=x+2 Substitution Please show your work Need ASAP

erik [133]

Answer:

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

Step-by-step explanation:

we have

$y=-x^{2} +2x+10$ ----> equation A

$y=x+2$ ----> equation B

Solve the system by substitution

substitute equation B in equation A

$x+2=-x^{2} +2x+10$

solve for x

$-x^{2} +2x+10-x-2=0$

$-x^{2} +x+8=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

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

in this problem we have

$-x^{2} +x+8=0$

$a=-1\\b=1\\c=8$

substitute in the formula

$x=\frac{-1\pm\sqrt{1^{2}-4(-1)(8)}} {2(-1)}$

$x=\frac{-1\pm\sqrt{33}} {-2}$

$x=\frac{-1+\sqrt{33}} {-2}$ -----> $x=\frac{1-\sqrt{33}} {2}$

$x=\frac{-1-\sqrt{33}} {-2}$ -----> $x=\frac{1+\sqrt{33}} {2}$

Find the values of y

For $x=\frac{1-\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1-\sqrt{33}} {2}+2$ ----> $y=\frac{5-\sqrt{33}} {2}$

For $x=\frac{1+\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1+\sqrt{33}} {2}+2$ ----> $y=\frac{5+\sqrt{33}} {2}$

therefore

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

5 0

3 years ago

A vending machine that serves coffee pours a varying amount of liquid into a cup with varying mass. Assume

bonufazy [111]

Answer:400

Step-by-step explanation:

7 0

3 years ago

A 24-foot ladder is placed against a vertical wall of a building, with the bottom of the ladder standing on level ground 6 feet

andrezito [222]

Answer:

23.24 feet

Step-by-step explanation:

Use the pythagorean theorem: a² + b² = c², where a and b are legs of the right triangle and c is the hypotenuse.

In this situation, the ladder is the hypotenuse of the triangle, and the distance from the base of the building is the long leg.

Plug in the ladder length as c and plug in the distance from the base of the building as a:

a² + b² = c²

(6²) + b² = (24)²

36 + b² = 576

b² = 540

b = 23.24

So, the ladder reaches approximately 23.24 feet up the wall

5 0

3 years ago

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