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3 years ago

8

A drilling machine can drill 12 holes in 6 minutes. If the machine drills holes at a constant speed, it can drill ______ holes i

n 25 minutes.

2 answers:

Nataly_w [17]3 years ago

5 0

50 might be the answer because 12÷6=2 so I just multiplied 15×2 and that equaled my answer, 50

eimsori [14]3 years ago

4 0

Answer:

50

Step-by-step explanation:

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Ms. Hernández is taking her children and their friends to the movies. She will pay $10 for her adult ticket and $7 for each chil

stiv31 [10]

(40-10)/7

The maximum amount of money she wants to spend is $40
Her adult ticket costs $10
She will pay for the children’s tickets with the remaining money
Each child’s ticket costs $7
To find the number of children she can take we need to use this expression (40-10)/7
Mrs Herñàndez can take 4 children

8 0

2 years ago

Read 2 more answers

ASAP BRAINLIEST if correct. Am i correct?

Minchanka [31]

Answer:

C complex

Step-by-step explanation:

a+ bi

a is the real part and bi is the imaginary part

Together they make a complex number

3 0

3 years ago

How do you write 60% as a fraction in simplest form?

saveliy_v [14]

60/100
6/10
3/5

Hope this helps :)

6 0

4 years ago

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Does anyone have any idea how to solve this problem?

Katena32 [7]

Answer:

$1350

Step-by-step explanation:

(150 x 30) + (300 x 45) = 18000

18000 x 0.075 = 1350

7 0

3 years ago

Simplify. Your answer should contain only positive exponents with no fractional exponents in the denominator.

mart [117]

Answer:

$\dfrac{x^{\frac{2}{3}}y^{\frac{1}{4}}}{4y^{2}}$ .

Step-by-step explanation:

The given expression is

$\dfrac{3y^{\frac{1}{4}}}{4x^{-\frac{2}{3}}y^{\frac{3}{2}}\cdot 3y^{\frac{1}{2}}}$

We need to simplify the expression such that answer should contain only positive exponents with no fractional exponents in the denominator.

Using properties of exponents, we get

$\dfrac{3}{4\cdot 3}\cdot \dfrac{y^{\frac{1}{4}}}{x^{-\frac{2}{3}}y^{\frac{3}{2}+\frac{1}{2}}}$ $[\because a^ma^n=a^{m+n}]$

$\dfrac{1}{4}\cdot \dfrac{y^{\frac{1}{4}}}{x^{-\frac{2}{3}}y^{2}}$

$\dfrac{1}{4}\cdot \dfrac{x^{\frac{2}{3}}y^{\frac{1}{4}}}{y^{2}}$ $[\because a^{-n}=\dfrac{1}{a^n}]$

$\dfrac{x^{\frac{2}{3}}y^{\frac{1}{4}}}{4y^{2}}$

We can not simplify further because on further simplification we get negative exponents in numerator or fractional exponents in the denominator.

Therefore, the required expression is $\dfrac{x^{\frac{2}{3}}y^{\frac{1}{4}}}{4y^{2}}$ .

5 0

3 years ago