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3 years ago

Please need help asap!!

Mathematics

1 answer:

worty [1.4K]3 years ago

6 0

A property of rectangles is that opposite sides are congruent. In this case, since QT and RS are opposite sides, they are equal to each other. With this knowledge, you can form the equation $50=4x+8$ to solve for x.

Firstly, subtract both sides by 8: $42=4x$

Next, divide both sides by 4 and your answer will be $10.5=x$ , or the third option.

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A square pyramid has base area 64 square centimeters. The surface area of this pyramid is 224 square centimeters. Find the heigh

dolphi86 [110]

Answer:

$h \approx 12.7\,cm$

Step-by-step explanation:

The side of the square base is:

$l = \sqrt{64\,cm^{2}}$

$l = 8\,cm$

The formula for the surface area of the pyramid is:

$A_{s} = \frac{1}{2}\cdot (3\cdot l)\cdot \sqrt{0.25\cdot l^{2}+h^{2}} + l^{2}$

The height is cleared in the previous expression:

$A_{s} - l^{2} = \frac{3}{2}\cdot l\cdot \sqrt{0.25\cdot l^{2}+h^{2}}$

$\frac{2}{3}\cdot \frac{A_{s}-l^{2}}{l} = \sqrt{0.25\cdot l^{2} + h^{2}}$

$\frac{4}{9}\cdot \left(\frac{A_{s}-l^{2}}{l} \right)^{2} = \frac{1}{4} \cdot l^{2} + h^{2}$

$h = \sqrt{\frac{4}{9}\cdot \left(\frac{A_{s}-l^{2}}{l} \right)^{2} - \frac{1}{4}\cdot l^{2}}$

$h = \sqrt{\frac{4}{9}\cdot \left[\frac{224\,cm^{2}-(8\,cm)^{2}}{8\,cm}\right]^{2}-\frac{1}{4}\cdot (8\,cm)^{2} }$

$h \approx 12.7\,cm$

6 0

3 years ago

Find the interquartile range for each data set.Round your answers to the nearest tenth. 19,13,15,26,11,15,17,5,2,0,2,2,4,3,4,6

Fed [463]

Answer:

12.5

Step-by-step explanation:

have a good day! :)

5 0

3 years ago

WILL MARK BRAINLIEST! WILL GIVE 57 POINTSSSS! NO FUNNY ANSWERS I WILL REPORT YOU UNTIL YOU GET BANNED.

Step2247 [10]

Answer:

-5

-3

-2

Rule: Subtract 2 from the input value

Step-by-step explanation:

Examine the pattern of the input and the corresponding output for the two given ordered pairs:

Input -2 → Output -4

Output - Input = -4 - (-2) = -2

Input 1 → Output -1

Output - Input = -1 - 1 = -2

Therefore to get the output value (y-value), we subtract 2 from the input value (x-value).

$\begin{array}{| c | c |}\cline{1-2} x & y \\\cline{1-2} -3 & -5 \\\cline{1-2} -2 & -4 \\\cline{1-2} -1 & -3 \\\cline{1-2} 0 & -2\\\cline{1-2} 1 & -1 \\\cline{1-2}\end{array}$