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3 years ago

Chen is checking a division problem by doing 152x4 equal to 608 plus 2 which is 610 what problem is chen checking?

Mathematics

2 answers:

Marizza181 [45]3 years ago

8 0

610 divided by 4 because when you divide that you will get 152

Ostrovityanka [42]3 years ago

8 0

Answer:

$610\div 4$

Step-by-step explanation:

In a division problem,

Dividend ÷ Divisor

We check the answer by the formula,

Dividend = quotient × divisor + remainder,

Here, the verification equation,

610 = 152 × 4 + 2,

By comparing,

We get, dividend = 610, divisor = 4,

The required problem would be,

$610\div 4$

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10 points

marishachu [46]

Answer:

150+12x=630

Step-by-step explanation:

The x represents the amount of months and it is multiplied by 12 because each month costs 12$. 150 have to be paid up front and this added to the monthly fee is equal to 630$

3 0

3 years ago

A juggler tosses a ball into the air . The balls height, h and time t seconds can be represented by the equation h(t)= -16t^2+40

malfutka [58]

PART A

The given equation is

$h(t) = - 16 {t}^{2} + 40t + 4$

In order to find the maximum height, we write the function in the vertex form.

We factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + 4$

We add and subtract

$- 16(- \frac{5}{4} )^{2}$

to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + - 16( - \frac{5}{4})^{2} - -16( - \frac{5}{4})^{2} + 4$

We again factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4})^{2} ) - -16( - \frac{5}{4})^{2} + 4$

This implies that,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4}) ^{2} ) + 16( \frac{25}{16}) + 4$

The quadratic trinomial above is a perfect square.

$h(t) = - 16 ( t- \frac{5}{4}) ^{2} +25+ 4$

This finally simplifies to,

$h(t) = - 16 ( t- \frac{5}{4}) ^{2} +29$

The vertex of this function is

$V( \frac{5}{4} ,29)$

The y-value of the vertex is the maximum value.

Therefore the maximum value is,

$29$

PART B

When the ball hits the ground,

$h(t) = 0$

This implies that,

$- 16 ( t- \frac{5}{4}) ^{2} +29 = 0$

We add -29 to both sides to get,

$- 16 ( t- \frac{5}{4}) ^{2} = - 29$

This implies that,

$( t- \frac{5}{4}) ^{2} = \frac{29}{16}$

$t- \frac{5}{4} = \pm \sqrt{ \frac{29}{16} }$

$t = \frac{5}{4} \pm \frac{ \sqrt{29} }{4}$

$t = \frac{ 5 + \sqrt{29} }{4} = 2.60$

or

$t = \frac{ 5 - \sqrt{29} }{4} = - 0.10$

Since time cannot be negative, we discard the negative value and pick,

$t = 2.60s$

8 0

3 years ago

1.325 to the nearest hundreths

12345 [234]

Its 1.33 because you round up if its 5 or greater

4 0

3 years ago

Read 2 more answers

~**Will mark brainliest **~<br> For the correct answers to all three questions

aleksandr82 [10.1K]

$\dfrac{42}{65}\cdot\dfrac{25}{36}\cdot\dfrac{26}{49}=\dfrac{2\cdot3\cdot7}{5\cdot13}\cdot\dfrac{5\cdot5}{2\cdot2\cdot3\cdot3}\cdot\dfrac{2\cdot13}{7\cdot7}=\dfrac{7}{13}\cdot\dfrac{5}{2\cdot3}\cdot\dfrac{2\cdot13}{7\cdot7}=\\\\\\=\dfrac{1}{1}\cdot\dfrac{5}{2\cdot3}\cdot\dfrac{2}{7}=\dfrac{5}{3}\cdot\dfrac{1}{7}=\dfrac{5}{21}$

$\dfrac{21}{32}\cdot\dfrac{39}{120}\cdot\dfrac{40}{65}=\dfrac{21}{32}\cdot\dfrac{3\cdot13}{2\cdot2\cdot2\cdot3\cdot5}\cdot\dfrac{2\cdot2\cdot2\cdot5}{5\cdot13}=\\\\\\=\dfrac{3\cdot7}{32}\cdot\dfrac{13}{2\cdot2\cdot2\cdot5}\cdot\dfrac{2\cdot2\cdot2}{13}=\dfrac{21}{32}\cdot\dfrac{1}{5}\cdot\dfrac{1}{1}=\dfrac{21}{160}$

$\dfrac{15}{90}\cdot\dfrac{36}{75}\cdot\dfrac{27}{42}=\dfrac{3\cdot5}{2\cdot3\cdot3\cdot5}\cdot\dfrac{2\cdot2\cdot3\cdot3}{3\cdot5\cdot5}\cdot\dfrac{3\cdot3\cdot3}{2\cdot3\cdot7}=\\\\\\=\dfrac{1}{2\cdot3}\cdot\dfrac{2\cdot2\cdot3}{5\cdot5}\cdot\dfrac{3\cdot3}{2\cdot7}=\dfrac{1}{1}\cdot\dfrac{2}{5\cdot5}\cdot\dfrac{3\cdot3}{2\cdot7}=\dfrac{1}{25}\cdot\dfrac{9}{7}=\dfrac{9}{175}$

6 0

3 years ago

The diagram below shows two parallel lines, m and n, cut by a transversal, k. Angles A, B, and C are shown in the diagram

puteri [66]

9514 1404 393

Answer:

Step-by-step explanation:

Angles A and C are vertical angles; angles B and C are alternate interior angles. Only line 3 of the proof is in error.

The applicable description is found in choice B.

8 0

3 years ago