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dangina [55]
3 years ago
11

Problem: -7(2)+y=60, so where do I start?

Mathematics
2 answers:
aev [14]3 years ago
4 0
-7(2)+y=60
-14+y=60
y=60+14
Y=74
VARVARA [1.3K]3 years ago
4 0
First simplify the left side of the equation:
-7(2)+y=60 -> -14+y=60
then you want to get y by it's self:
-14+y=60
-14 -14
------------------=> y=60-14
now simplify the right side of the equation:
y= 46
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A website received 2500 registered members in July. And the number of registered members in August decreased to 2245. What was t
quester [9]

Number of registrations a website received during July = 2500

Number of registrations a website received during August = 2245

Number by which the registrations decreased :

=\tt 2500 - 2245

=\tt 255

Thus, the registrations decreased by 255.

Let x be the perrcentage of decrease.

Which means :

=\tt 255 =  \frac{x}{100}  \: of \: 2500

= \tt255 =  \frac{x \times 2500}{100}

=\tt 255 =  \frac{2500x}{100}

= \tt255 \times 100  = 2500x

= \tt25500 = 2500x

=\tt  \frac{25000}{2500}  = x

\color{plum} =\tt  \bold{10.2\%}

▪︎Thus, the percentage of decrease in the number of registered members = 10.2%

5 0
3 years ago
Im so lost, please help me asap
almond37 [142]

Answer:

it's so simple ....

x + 141 = 135

x = 135 - 141

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it means ur answer is a

8 0
3 years ago
Marcos purchased a sail boat for $56,900. He anticipates each year the boat will decrease in value by 8.1% from the previous yea
Varvara68 [4.7K]

Answer: 56,900\left( 0.919\right)^x

Step-by-step explanation:

Given

Marcos purchased a sailboat for \$56,900

Each year the boat will decrease in value by 8.1\%

After 1 year it is

\Rightarrow 56,900-56,900\times 0.081\\\Rightarrow 56,900\left( 1-0.081\right)\\

after another year it becomes

\Rightarrow 56,900\left( 1-0.081\right)-56,900\left( 1-0.081\right)\times 0.081\\\Rightarrow 56,900\left( 1-0.081\right)\cdot \left( 1-0.081\right)\\\Rightarrow 56,900\left( 1-0.081\right)^2

After x years it is

\Rightarrow 56,900\left( 1-0.081\right)^x

\Rightarrow 56,900\left( 0.919\right)^x

5 0
3 years ago
An object is dropped from a small plane. As the object falls, its distance, d, above the ground after t seconds, is given by the
DedPeter [7]

<em><u>The inequality can be used to find the interval of time taken by the object to reach the height  greater than 300 feet above the ground is:</u></em>

d = -16t^2 + 1000

<em><u>Solution:</u></em>

<em><u>The object falls, its distance, d, above the ground after t seconds, is given by the  formula:</u></em>

d = -16t^2 + 1000

To find the time interval in which the object is at a height greater than 300 ft

Frame a inequality,

-16t^2 + 1000 > 300

Solve the inequality

Subtract 1000 from both sides

-16t^2 + 1000 - 1000 > 300 - 1000\\\\-16t^2 > -700

16t^2 < 700\\\\Divide\ both\ sides\ by\ 16\\\\t^2 < \frac{700}{16}\\\\Take\ square\ root\ on\ both\ sides\\\\t < \sqrt{\frac{700}{16}}\\\\t < \pm 6.61

Time cannot be negative

Therefore,

t < 6.61

And the inequality used  is: -16t^2 + 1000>300

6 0
3 years ago
A well mixed tank with a capacity of 1500 gals originally contains 1000 gallons of freshwater. One pipe containing 1/2 lb of sal
Illusion [34]

Answer:

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Water flows out at 5 gal/min, therefore in x minute the amount of water in the tank = 1000 + 10x - 5x = 1000 + 5x

The tank begins to overflow when it is full (has reached 1500 gallons). Therefore:

1500 = 1000 + 5x

5x = 1500 - 1000

5x = 500

x = 100 minutes.

1/2 lb salt per gallon flows into the tank at 4 gal/min and 1/3 lb of salt is flowing in at 6 gal/min, in 100 min the amount of salt that entered the tank = 4 gal/min × 100 min × 1/2 lb/gal + 6 gal/min × 100 min × 1/3 lb/gal= 400 lb

Therefore the amount of salt is in the tank when it is about to overflow = 400 lb of salt

7 0
3 years ago
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