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2 years ago

Solve 18 > 12 + x. Graph the solution.

Mathematics

1 answer:

telo118 [61]2 years ago

3 0

Answer:

I have attached a screenshot of your inequality graphed. Credit to Desmos.

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Y’all. I need you to help me

Sergio039 [100]

Answer:

Step-by-step explanation:

PQ/QR = ST/TR

17.5/QR = 7/4

QR = (17.5 x 4)/7

QR = 70/7

QR = 10

3 0

3 years ago

What is the slope of a linear that is perpendicular to the line y=-1/2x+5?

Vlada [557]

Step-by-step explanation:

Equation of given line is:

$y = - \frac{1}{2} x + 5 \\ \\ equating \: it \: with \: y =m_1x + c \\ we \: find \: \\ m_1 = - \frac{1}{2} \\ let \: m_2 \: be \: the \: slope \: of \: required \: line \\ \\ \therefore \: m_1 \times m_2 = - 1( \because \: lines \: are \: \perp) \\ \\ \therefore \: - \frac{1}{2} \times m_2 = - 1\\ \\ \therefore \: m_2 = 2 \\$

Thus, the slope of required line is 2.

6 0

3 years ago

You have quarters and nickels saved in a piggy bank. There is a total of $3.45 in the bank. If you have 70 cents in nickels, how

Lerok [7]

Answer:If I have quarters and nickels saved in my piggy bank I know that 9 quarters is $2.25 and i got 28 nickels that equal up to $3.45

Step-by-step explanation:

4 0

3 years ago

A greyhound ran 425 yards in<br> 25 seconds. How fast was the<br> greyhound running?

Afina-wow [57]

17 yards per second

5 0

3 years ago

Read 2 more answers

A certain bridge arch is in the shape of half an ellipse 106 feet wide and 33.9 feet high. At what horizontal distance from the

nata0808 [166]

Answer:

The horizontal distance from the center is 49.3883 feet

Step-by-step explanation:

The equation of an ellipse is equal to:

$\frac{x^2}{a^{2} } +\frac{y^2}{b^{2} } =1$

Where a is the half of the wide, b is the high of the ellipse, x is the horizontal distance from the center and y is the height of the ellipse at that distance.

Then, replacing a by 106/2 and b by 33.9, we get:

$\frac{x^2}{53^{2} } +\frac{y^2}{33.9^{2} } =1\\\frac{x^2}{2809} +\frac{y^2}{1149.21} =1$

Therefore, the horizontal distances from the center of the arch where the height is equal to 12.3 feet is calculated replacing y by 12.3 and solving for x as:

$\frac{x^2}{2809} +\frac{y^2}{1149.21} =1\\\frac{x^2}{2809} +\frac{12.3^2}{1149.21} =1\\\\\frac{x^2}{2809}=1-\frac{12.3^2}{1149.21}\\\\x^{2} =2809(0.8684)\\x=\sqrt{2809(0.8684)}\\x=49.3883$

So, the horizontal distance from the center is 49.3883 feet

8 0

3 years ago