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Andrei [34K]
3 years ago
5

Patricia scored 80% on a math test. She missed 4 problems. How many problems were on the test?

Mathematics
2 answers:
valina [46]3 years ago
7 0

Answer: There are 20 problems on the test.

Step-by-step explanation:

Since we have given that

Patricia scored 80% on a math test.

She missed 4 problems.

So, 20% of the math test is equal to 4 problems.

Let the total number of problems on the test be x

According to question,

\frac{20}{100}\times x=4\\\\x=\frac{100}{20}\times 4\\\\x=20

Hence, there are 20 problems on the test.

arsen [322]3 years ago
6 0
There would be exactly twenty problems on the test for her to be able to get 80%.
<span>You know it can't be ten because then that would make it 60% not 80% but if you add the first number of the percentage (8) and multiply it by two it will get you 16 then if you add 4 to that that would make it 20 and 16/20 equals 80%</span>
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()) 91 is the quotient of c and 283​
Crazy boy [7]

Hey there,

The word problem is as follows:

  • 91 is the quotient of c and 283

From the word choice you know that this is a division problem therefore in numbers this would become 283 ÷ c = 91.

Now that we know that in order to find c we would have to divide 283 by 91 which is another way to write that equation.

So...

  • 283 ÷ 91 would be 3 \frac{10}{91}

Hope I helped,

Amna

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4 years ago
Determine whether each first-order differential equation is separable, linear, both, or neither. 1. ????y????x+????xy=x2y2 2. y+
Mkey [24]

Answer:

a) Linear

b) Linear

c) Linear

d) Neither

See explanation below.

Step-by-step explanation:

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For this case the differential equation have the following general form:

y' +p(x) y = q(x) y^n

Where p(x) =e^x and q(x) = x^2 and since n>1 we can see that is a linear differential equation.

b) y + sin x = x^3 y'

We can rewrite the following equation on this way:

y' -\frac{1}{x^3} y= \frac{sin (x)}{x^3}

For this case the differential equation have the following general form:

y' +p(x) y = q(x) y^n

Where p(x) =-\frac{1}{x^3} and q(x) = \frac{sin(x)}{x^3} and since n=0 we can see that is a linear differential equation.

c) ln x -x^2 y =xy'

For this case we can write the differential equation on this way:

y' +xy = \frac{ln(x)}{x}

For this case the differential equation have the following general form:

y' +p(x) y = q(x) y^n

Where p(x) =x and q(x) = \frac{ln(x)}{x} and since n=0 we can see that is a linear differential equation.

d) \frac{dy}{dx} + cos y = tan x

For this case we can't express the differential equation in terms:

y' +p(x) y = q(x) y^n

So the is not linear, and since we can separate the variables in order to integrate is not separable. So then the answer for this one is neither.

4 0
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EleoNora [17]

Answer:

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Step-by-step explanation:

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6 0
4 years ago
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mojhsa [17]

Answer:

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Step-by-step explanation:

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