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4 years ago

One number is 5 times another number. The product of the two numbers is 245. Find the two numbers.

Mathematics

1 answer:

o-na [289]4 years ago

8 0

There are two equations that equal 245.

<span>5 x 49 or 7 x 35 equals 245. I hope this helps!</span>

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Simplify the expression and rewrite in rational exponent form.

Dahasolnce [82]

Answer:

$4 x^{\frac{11}{10}} \cdot y^{\frac{17}{3}}$

Step-by-step explanation:

The given expression: $4 \sqrt[5]{x^{3}} \cdot y^{4} \cdot \sqrt{x} \cdot \sqrt[3]{y^{5}}$

Step 1: Change radical to fractional exponent.

Formula for fractional exponent: $\sqrt[n]{a}=a^{\frac{1}{n}}$

The power to which the base is raised becomes the numerator and the root becomes the denominator.

$\Rightarrow 4 x^{\frac{3}{5}} \cdot y^{4} \cdot x^{\frac{1}{2}} \cdot y^{\frac{5}{3}}$

Step 2: Apply law of exponent for a product $a^{m} \times a^{n}=a^{m+n}$

Multiply powers with same base.

$\Rightarrow 4 x^{\frac{3}{5}+\frac{1}{2}} \cdot y^{4+\frac{5}{8}}$

Take LCM for the fractions in the power.

$\Rightarrow 4 x^{\frac{6}{10}+\frac{5}{10}} \cdot y^{\frac{12}{3}+\frac{5}{3}}$

$\Rightarrow 4 x^{\frac{11}{10}} \cdot y^{\frac{17}{3}}$

Hence the simplified form of $4 \sqrt[5]{x^{3}} \cdot y^{4} \cdot \sqrt{x} \cdot \sqrt[3]{y^{5}} \text { is } 4 x^{\frac{11}{10}} \cdot y^{\frac{17}{3}}$ .

5 0

3 years ago

Since the expression (x-4)^2+(y+1)^2 evaluated at (3,6) is

stiks02 [169]

$\bf (x-4)^2+(y+1)^2\qquad (3,6)~~ \begin{cases} x = 3\\ y = 6 \end{cases}\implies (3-4)^2+(6+1)^2 \\\\\\ (-1)^2+7^2\implies 1+49\implies 50$

5 0

3 years ago

Y+6=-5(x-6) to slope-intercept form

kirill [66]

Answer:

y=-5x-36

Step-by-step explanation:

distribute that into:

y+6=-5x-30

subtract 6 and get the answer

5 0

3 years ago

Read 2 more answers

HELP. there are 45 coins consisting of nickels, dimes and quarter making a total of 7 dollars. the number of dimes exceeds the n

MA_775_DIABLO [31]

9514 1404 393

Answer:

Step-by-step explanation:

Let n, d, q represent the numbers of nickels, dimes, and quarters, respectively. Then we have ...

n + d + q = 45

5n +10d +25q = 700

n -d = -5

Using n=d -5, we can substitute into the first two equations:

(d -5) +d + q = 45

2d +q = 50 . . . . add 5, collect terms [eq4]

5(d -5) +10d +25q = 700

15d +25q = 725 . . . . . . . . . . add 25, collect terms [eq5]

Multiplying [eq4] equation by 3 and subtracting that from 2/5 of [eq5], we have ...

(2/5)(15d +25q) -3(2d +q) = (2/5)(725) -3(50)

6d +10q -6d -3q = 290 -150

7q = 140 . . . . . . simplify

q = 20 . . . . . . . . divide by 7

There are 20 quarters.

8 0

3 years ago

Toy cars cost $2. If Anna bought 4 toy cars and 5 toy airplanes and spent $38, how

Y_Kistochka [10]

Answer: A toy airplane costs $6

Step-by-step explanation: Because 4 toy cars cost $8 , then there are $30 for 5 toy planes. One toy plane cost 30/5 = $6

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3 years ago