Megan and Suzanne each have a plant. They track the growth of their plants for four weeks.

Compare 9 × (11 – 4) and 3 × (11 – 4). A. 9 × (11 – 4) is three times as much as 3 × (11 – 4). B. 9 × (11 – 4) is nine times as

liubo4ka [24]

Hello there!

So there are two separate problems: 9 * (11-4) and 3 * (11-4). To find the value for each expression, use order of operations. We'll solve what's in the parenthesis before multiplying that value by the other number. 11-4 is 7. 7 * 9 is 63. 11-4 is 7. 7 * 3 is 21. You can divide 63 by 21 to get 3. This means that 9 * (11 - 4) is three times as much as 3 *(11 - 4). Even just by looking at the problem, the expression in the parenthesis is the same thing, but the front numbers are different. 9 is three times as much as 3. The answer is A.

3 0

3 years ago

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A factory is able to manufacture parts for a ceiling fan base on the data shown in the graph below how many parts can the factor

just olya [345]

Answer:

Step-by-step explanation:

Yo to be honest I’m looking for the answer too. Just ask your math teacher bruh.

7 0

3 years ago

Sean earns $4.50 per hour more than his trainee, Tyrone. During one 8 hour day, they earn a total of $284. How much does Tyrone

LenKa [72]

I think it is 36 well to what ik :)

7 0

4 years ago

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What is the measurement of the longest line segment in a right rectangular prism that is 26 inches long, 2 inches wide, and 2 in

EastWind [94]

Answer:

$6\sqrt{19} \approx 26.153$ inches.

Step-by-step explanation:

The longest line segment in a right rectangular prism is the diagonal that connects two opposite vertices. On the first diagram attached, the green line segment connecting A and G is one such diagonals. The goal is to find the length of segment $\mathsf{AG}$ .

In this diagram (not to scale,) $\mathsf{AB} = 26$ (length of prism,) $\mathsf{AC} = 2$ (width of prism,) $\mathsf{AE} = 2$ (height of prism.)

Pythagorean Theorem can help find the length of $\mathsf{AG}$ , one of the longest line segments in this prism. However, note that this theorem is intended for right triangles in 2D, not the diagonal in a 3D prism. The workaround is to simply apply this theorem on two different right triangles.

Start by finding the length of line segment $\mathsf{AD}$ . That's the black dotted line in the diagram. In right triangle $\triangle\mathsf{ABD}$ (second diagram,)

Segment $\mathsf{AD}$ is the hypotenuse.
One of the legs of $\triangle\mathsf{ABD}$ is $\mathsf{AB}$ . The length of $\mathsf{AB}$ is $26$ , same as the length of this prism.
Segment $\mathsf{BD}$ is the other leg of this triangle. The length of $\mathsf{BD}$ is $2$ , same as the width of this prism.

Apply the Pythagorean Theorem to right triangle $\triangle\mathsf{ABD}$ to find the length of $\mathsf{AB}$ , the hypotenuse of this triangle:

$\mathsf{AD} = \sqrt{\mathsf{AB}^2 + \mathsf{BD}^2} = \sqrt{26^2 + 2^2}$ .

Consider right triangle $\triangle \mathsf{ADG}$ (third diagram.) In this triangle,

Segment $\mathsf{AG}$ is the hypotenuse, while
$\mathsf{AD}$ and $\mathsf{DG}$ are the two legs.

$\mathsf{AD} = \sqrt{26^2 + 2^2}$ . The length of segment $\mathsf{DG}$ is the same as the height of the rectangular prism, $2$ (inches.) Apply the Pythagorean Theorem to right triangle $\triangle \mathsf{ADG}$ to find the length of the hypotenuse $\mathsf{AG}$ :

$\begin{aligned}\mathsf{AG} &= \sqrt{\mathsf{AD}^2 + \mathsf{GD}^2} \\ &= \sqrt{\left(\sqrt{26^2 + 2^2}\right)^2 + 2^2}\\ &= \sqrt{\left(26^2 + 2^2\right) + 2^2} \\&= 6\sqrt{19} \\&\approx 26.153\end{aligned}$ .

Hence, the length of the longest line segment in this prism is $6\sqrt{19} \approx 26.153$ inches.

5 0

3 years ago

Solve 1 over 36 = 6x−3. x = −5 x = negative 3 over 2 x = 7 over 2 x = 1

vlabodo [156]

6*(-5)-3 = -33
6*(-3/2)-3 = -12
6*(7/2)-3 = 18
6*(1)-3=3

3 0

3 years ago

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