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3 years ago

What is the slope of the line that passes through the points (8,7) and (4,5)?

Mathematics

2 answers:

musickatia [10]3 years ago

6 0

Answer:

The slope is $\frac{1}{2}$

Step-by-step explanation:

The slope formula is $\frac{y2-y1}{x2-x1}$

(8, 7) is (x1, y1)

(4, 5) is (x2, y2)

Plug in the values:

$\frac{5-7}{4-8}$

solve:

$\frac{-2}{-4}$

$\frac{1}{2}$

The slope is $\frac{1}{2}$

Varvara68 [4.7K]3 years ago

3 0

Answer:

1/2

Step-by-step explanation:

Y2-y1/x2-x1

-2/-4

1/2

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2 years ago

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Savannah has a principle of 15000 with a 4.25% and saves 1,275 in interest. How long is she saving for

Nostrana [21]

Answer:

5.73 years

Step-by-step explanation:

Considering she saved 4.25% in each year of the amount $15000.

So, we have to calculate in what time her saving reaches $1275, if she save each year.

it becomes question of simple interest in which

SI = $1275, Principal P= $5000, time T, rate R = 4.25%

we know SI = PRT/100

1275 = 5000×4.25×T/100

⇒T = 1275×100/(5000×4.25)

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2 years ago

What is the value(s) of x?

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7 0

3 years ago

An oak dresser has a volume of 0.1cubic meters and a mass of 90kilograms. Calculate its density.

Mars2501 [29]

Answer:

900kg/ $m^{3}$

Step-by-step explanation:

Formula: d = m/v

Given: m = 90, v = 0.1

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2 years ago

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Michelle's father deposited $1,000 into a savings account when she was born. The annual interest rate is

prohojiy [21]

Answer:

Michelle's age = 16

Step-by-step explanation:

As we know,

Amount = $P (1 + r)^{t}$ where

P = Principal amount

r = rate of interest

t = time

Given ,

P = $1000

r = 4.5 % = $\frac{4.5}{100} = 0.045$

Now,

Given that Amount reaches more than $2000

⇒ $P (1 + r)^{t}$ > 2000

⇒ $1000(1 + 0.045)^{t}$ > 2000

⇒ $(1.045)^{t}$ > 2

Now,

we put the value of t from 0 , 1, 2, ..... untill the above equation does not satisfy

Now,

for t = 0

$(1.045)^{0} = 1 \ngeq 2$

for t = 1

$(1.045)^{1} = 1.045 \ngeq 2$

for t = 2

$(1.045)^{2} = 1.092 \ngeq 2$

for t = 3

$(1.045)^{3} = 1.14 \ngeq 2$

for t = 4

$(1.045)^{4} = 1.19 \ngeq 2$

for t = 5

$(1.045)^{5} = 1.25 \ngeq 2$

for t = 6

$(1.045)^{6} = 1.30 \ngeq 2$

for t = 7

$(1.045)^{7} = 1.36 \ngeq 2$

for t = 8

$(1.045)^{8} = 1.42 \ngeq 2$

for t = 9

$(1.045)^{9} = 1.48 \ngeq 2$

for t = 10

$(1.045)^{10} = 1.55 \ngeq 2$

for t = 11

$(1.045)^{11} = 1.62 \ngeq 2$

for t = 12

$(1.045)^{12} = 1.69 \ngeq 2$

for t = 13

$(1.045)^{13} = 1.77 \ngeq 2$

for t = 14

$(1.045)^{14} = 1.85 \ngeq 2$

for t = 15

$(1.045)^{15} = 1.93 \ngeq 2$

for t = 16

$(1.045)^{16} = 2.02 > 2$

It satisfies at t= 16

∴ we get

Michelle's age = 16

4 0

3 years ago