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Anton [14]
3 years ago
9

as part of a summer job, you stack crates. the crates have the same length and width but have heights of 1 or 2 feet. using a fo

rklift, you can stack the crates 8 feet high. if you stacked 3 of the 2-foot crates how many of the 1-foot crates were in the stack?
Mathematics
1 answer:
Pavlova-9 [17]3 years ago
4 0

Answer:

1 of the feet is 2 of the  feet

Step-by-step explanation:

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C and ​ ∠D ​ are vertical angles with m∠C=−x+26 and m∠D=2x−10 . What is m∠D ? Enter your answer in the box. ​ °
ANEK [815]
Angle c and angle d are congruent. Congruent means they have the same the measure, or they have EQUAL measures. You set the two values equal to each other and solve for x.
X+26=2x-10 subtract x from both sides
26=x-10 add 10 to both sides
36=x

Now substitute x back in for angle d.
2(36)-10 = 62
4 0
3 years ago
Find in the missing values to make the equations true.
Vikki [24]

Answer: 21

12

Log8(8 is under the log) the answer is 7

Step-by-step explanation:

8 0
3 years ago
0.003 is 1/10 of ____
34kurt
.0003, .04, .02, 2 now i need to type more do i cam be more than 20 characters
8 0
3 years ago
I need help ASAP!!!!!!!
solmaris [256]

Answer: i think the answer is d or c im not to sure

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
A circle passes through points A(7,4), B(10,6), C(12,3). Show that AC must be the diameter of the circle.
Artist 52 [7]

so we have three points, A, B and C, if indeed AC is the diameter of the circle, then half the distance of AC is its radius, and the midpoint of AC is the center of the circle, morever, since B is also on the circle, the distance from B to the center must be the same radius distance.

in short, half the distance of AC must be equals to the distance of B to the midpoint of AC, if indeed AC is the diameter.

\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{7}~,~\stackrel{y_1}{4})\qquad C(\stackrel{x_2}{12}~,~\stackrel{y_2}{3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{12+7}{2}~~,~~\cfrac{3+4}{2} \right)\implies \left( \cfrac{19}{2}~~,~~\cfrac{7}{2} \right)=M\impliedby \textit{center of the circle}

now, let's check the distance from say A to the center, and check the distance of B to the center, if it's indeed the center, they'll be the same and thus AC its diameter.

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{7}~,~\stackrel{y_1}{4})\qquad M(\stackrel{x_2}{\frac{19}{2}}~,~\stackrel{y_2}{\frac{7}{2}})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ AM=\sqrt{\left( \frac{19}{2}-7 \right)^2+\left( \frac{7}{2}-4 \right)^2} \\\\\\ AM=\sqrt{\left( \frac{5}{2}\right)^2+\left( -\frac{1}{2} \right)^2}\implies \boxed{AM\approx 2.549509756796392} \\\\[-0.35em] ~\dotfill

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ B(\stackrel{x_1}{10}~,~\stackrel{y_1}{6})\qquad M(\stackrel{x_2}{\frac{19}{2}}~,~\stackrel{y_2}{\frac{7}{2}}) \\\\\\ BM=\sqrt{\left( \frac{19}{2}-10 \right)^2+\left( \frac{7}{2}-6 \right)^2} \\\\\\ BM=\sqrt{\left( -\frac{1}{2}\right)^2+\left( -\frac{5}{2} \right)^2}\implies \boxed{BM\approx 2.549509756796392}

6 0
3 years ago
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