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2 years ago

Can someone help me with this

Mathematics

1 answer:

Mazyrski [523]2 years ago

4 0

Answer:

x = 3.6

Step-by-step explanation:

5.4 (1/3) = 1.8

2 (1.8) = 3.6

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Solve the following equation algebraically. Shiw your work. 17=-13-8x

maxonik [38]

17=13-8x

-13 . -13

4=8x

/8 . /8

x=1/2 OR 0.5

8 0

2 years ago

Johny Deposited 50000 rupees, Jaleel 40000 and Jaleel 20000 rupees to start a business together.They got 3300 rupees as a profit

Andrews [41]

Step-by-step explanation:

Step one:

Johny Deposited 50000 rupees,

Jaleel 40000 and

Jaleel 20000 rupees

total deposits = 110,000

They got 3300 rupees as a profit in a month

Johny's investment in percent

50000/110000= 0.45%

Jaleel's investment in percent

40000/110000= 0.36%

Jaleel's(2) investment in percent

50000/110000= 0.18%

They will get

Johny =0.45*33000= 1485 rupees

Jaleel= 0.36*3300= 1188 rupees

Jaleel(2)= 0.18*3300=594 rupees

7 0

2 years ago

If there are approximately 1.8 million cars in Vancouver and if each car travels an average of 17,500 kilometres in a year, how

Goryan [66]

Rates are used to measure a quantity over another.

The 1.8 million cars use $1.85 \times 10^{10}$ liters each year

Given

$Cars = 1.8\ million$

$d = 17500km$ --- distance

$r = 40miles/gallon$

First, we calculate the number (n) of gallons used by each car

$r = \frac{d}{n}$

Solve for n

$n = \frac{d}{r}$

So, we have:

$n = \frac{17500km}{40miles/gallon}$

$n = \frac{17500km}{40miles}gallon$

Convert miles to kilometers

$n = \frac{17500km}{40km \times 1.60934}gallon$

$n = \frac{17500km}{64.3738km}gallon$

$n = 271.85\ gallon$

The number of gallons (N), used by all the cars is:

$N = n \times cars$

$N = 271.85 \times 1.8\ million$

$N = 489330000$

Convert to liters

$N = 489330000 \times 3.78541L$

$N = 1852314675.3L$

In scientific notation to 2 decimal places, we have:

$N = 1.85 \times 10^{10}L$

Hence, the number of liters used is $1.85 \times 10^{10}$

Read more about distance and rates at:

brainly.com/question/24659604

4 0

2 years ago

The endpoints of EF are E(1,3) and F(-2,8). Find EF.

Phantasy [73]

Answer:

EF = $\sqrt{34}$ ≈ 5.83

Step-by-step explanation:

Calculate EF using the distance formula

d = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2 }$

with (x₁, y₁ ) = E(1, 3) and (x₂, y₂ ) = F(- 2, 8)

EF = $\sqrt{(-2-1)^2+(8-3)^2}$

= $\sqrt{(-3)^2+5^2}$

= $\sqrt{9+25}$

= $\sqrt{34}$ ≈ 5.83 ( to 2 dec. places )

8 0

2 years ago

On a coordinate plane, Point J is plot at (-4,6) and Point K is plotted at (8,-3). What is the distance, in units, between point

zhuklara [117]

Idk I tried this but I can't get it

4 0

3 years ago