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3 years ago

Can you help me find x? 37% of x is 44.03

Mathematics

1 answer:

balandron [24]3 years ago

5 0

Answer:

119

Step-by-step explanation:

37% = 44.03

1% = 44.03 ÷ 37 = 1.19

100% = 1.19 x 100 = 119 (100% is x)

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What is the difference between 5 hundreds and 3 tens?

Umnica [9.8K]

Answer:

5 tens and 3 hundreds = 50 + 300 = 350. That is basically asking what number goes 5-3-2. This is because 5 hundreds means 5 in the hundreds place, or 500. 3 tens is just saying 3 in the tens place, or 30.

Step-by-step explanation:

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3 years ago

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A grocery store offers a 10% discount on your entire purchase when you spend more than $50. You have $117 to spend on groceries

astra-53 [7]

Answer:

a≤80

Step-by-step explanation:

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3 years ago

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PLZZZZZZZ HELPP Which ordered pair is the solution to the system of equations? {y=2x−15} {4x+3y=−5} (5, −5) (4, −7) (0, 6) (7

Sophie [7]

solution is (4, - 7 )

given the equations

y = 2x - 15 → (1)

4x + 3y = - 5 → (2)

substitute y = 2x - 15 into (2)

4x + 3(2x - 15) = - 5

4x + 6x - 45 = - 5

10x - 45 = - 5 ( add 45 to both sides )

10x = 40 ( divide both sides by 10 )

x = 4

substitute x = 4 into (1) for corresponding value of y

y = (2 × 4 ) - 15 = 8 - 15 = - 7

solution is (4, - 7 )

8 0

3 years ago

Help I don’t understand at all

zloy xaker [14]

Answer:

$1)\ \ 4h^2-13h+6\\2)\ \ 7x^3y^2-x^2y+1\\3)\ \ -7n+2\\4)\ \ -8m+4$

Step-by-step explanation:

Simplify the expression by combining like terms. Remember, like terms have the same variable part, to simplify these terms, one performs operations between the coefficients. Please note that a variable with an exponent is not the same as a variable without the exponent. A term with no variable part is referred to as a constant, constants are like terms.

$2h^2-7h+2h^2-h+6+4h-9h$

$(2h^2+2h^2)+(-7h-h+4h-9h)+(6)$

$4h^2-13h+6$

Use a very similar method to solve this problem as used in the first. Please note that all of the rules mentioned in the first problem also apply to this problem; for that matter, the rules mentioned in the first problem generally apply to any pre-algebra problem.

$8x^3y^2-7x^2y+8x-4-x^3y^2+2x^2y+4x^2y-8x+5$

$(8x^3y^2-x^3y^2)+(-7x^2y+2x^2y+4x^2y)+(8x-8x)+(-4+5)$

$7x^3y^2-x^2y+1$

Use the same rules as applied in the first problem. Also, keep the distributive property in mind. In simple terms, the distributive property states the following ( $a(b+c)=(a)(b)+(a)(c)=ab+ac$ ). Also note, a term raised to an exponent is equal to the term times itself the number of times the exponent indicates. In the event that the term raised to an exponent is a constant, one can simplify it. Apply these properties here,