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3 years ago

Write an equation to represent the following statement.

Mathematics

1 answer:

hichkok12 [17]3 years ago

5 0

Answer:

j / 9 = 5

j = 45

Step-by-step explanation:

j / 9 = 5

Multiply each side by 9

j/9 * 9 = 5*9

j = 45

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Which relatives between x and y in the equation shows a proportional relationship?

tangare [24]

its B Step-by-step explanation:

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3 years ago

12.92 divided by 3.4 show work please

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3 years ago

An explosion causes debris to rise vertically with an initial speed of 120 feet per second. The formula h equals negative 16 t s

Novay_Z [31]

Answer:

The debris will be at a height of 56 ft when time is <u>0.5 s and 7 s.</u>

Step-by-step explanation:

Given:

Initial speed of debris is, $s=120\ ft/s$

The height 'h' of the debris above the ground is given as:

$h(t)=-16t^2+120t$

As per question, $h(t)=56\ ft$ . Therefore,

$56=-16t^2+120t$

Rewriting the above equation into a standard quadratic equation and solving for 't', we get:

$-16t^2+120t-56=0\\\textrm{Dividing by -8 throughout, we get}\\\frac{-16}{-8}t^2+\frac{120}{-8}t-\frac{56}{-8}=0\\2t^2-15t+7=0$

Using quadratic formula to solve for 't', we get:

$t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\t=\frac{-(-15)\pm \sqrt{(-15)^2-4(2)(7)}}{2(2)}\\\\t=\frac{15\pm \sqrt{225-56}}{4}\\\\t=\frac{15\pm\sqrt{169}}{4}\\\\t=\frac{15\pm 13}{4}\\\\t=\frac{15-13}{4}\ or\ t=\frac{15+13}{4}\\\\t=\frac{2}{4}\ or\ t=\frac{28}{4}\\\\t=0.5\ s\ or\ t=7\ s$

Therefore, the debris will reach a height of 56 ft twice.

When time $t=0.5\ s$ during the upward journey, the debris is at height of 56 ft.

Again after reaching maximum height, the debris falls back and at $t=7\ s$ , the height is 56 ft.

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3 years ago

Kamden is a mind-reading alien. He had each friend draw and look at a card, then he guessed whether the card had a star on it.

yulyashka [42]

The completed table will have values filled into it accordingly from left to right and the downwards.

<h3>What values are required to complete the table?</h3>

The concept of relative frequencies involves expressing the frequency of occurrence of events as a fraction of a whole frequency.

Hence, it follows that the blank space on the top-leftmost corner of the table can be filled with the value; 4/5 = 0.8.

For the top-rightmost blank space in the table, the required value is; 2/754 = 0.0027.

For the bottom-leftmost blank space, the required value to fill the table is; 1/5 = 0.2.

For the bottom-rightmost blank space, the required value is; 752/754 = 0.9973.