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algol13
1 year ago
6

A 6 inch by 10 inch rectangle is dilated by afactor of 3. What is the area of the dilatedrectangle?

Mathematics
1 answer:
qaws [65]1 year ago
5 0

Given;

There are given that the 6 inches by 10-inch rectangle.

Explanation:

dilated by a factor of three means that the two dimensions needed for the area are multiplied by 3.

So,

The width of the rectangular is:

width=6\times3=18

And,

The length of the rectangular is:

10\times3=30

Then,

The area of dilated rectangular is:

\begin{gathered} Area\text{ of dilated rectanular=}width\times\text{length} \\ =18\times30 \\ =540inch^2 \end{gathered}

Final answer:

Hence, the area of the dilated rectangle is 540 inch^2.

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Use the additive identity property to arrive at 6x=48.
andreev551 [17]

<h2>8</h2>

→ 6x = 48

→ x = 48/6

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HOPE THIS HELPS YOU

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7 0
3 years ago
The radius of a frisbee is 5 inches. What is the circumference of the frisbee in terms of (pi)?
vaieri [72.5K]

Answer:

10π inches

Step-by-step explanation:

C = 2πr

C = 2π(5)

C = 10π ≈ 31.42

4 0
3 years ago
An iron ball is bobbing up and down on the end of a spring. The maximum height of the ball is 50 centimeters and its minimum hei
ololo11 [35]

Answer:

Step-by-step explanation:

Let us examine all options:

(a)

Lets verify for t = 0:

h(t = 0) = 7

This isn't allowable as 14 cm is the lower limit for the ball's motion.

This is rejected.

(b)

Let's verify for t = 0:

h(t = 0) = 14

Since at t = 0 sec, the ball is at its minimum, after 2 seconds, it should be at its maximum.

But h(t = 2) = 14, which doesn't satisfy the condition.

Hence, this is rejected.

(c)

Now, let's see at what time instance, the ball is at minimum in this case:

h(t) = 14 = 18sin(πt) + 32

∴ sin(πt) = -1

∴ πt = 3π/2

∴ t = 3/2 seconds

Hence, after 2 seconds, i.e. at 3.5 seconds, the ball should be at its maximum.

h(t = 3.5) = 18sin(3.5π) + 32 = -18 + 32 = 14, which doesn't satisfy the condition.

Hence, this is rejected,

(d)

Now, let's see at what time instance, the ball is at minimum in this case:

h(t) = 14 = 18sin(πt/2) + 32

∴ sin(πt/2) = -1

∴ πt/2 = 3π/2

∴ t = 3 seconds

Hence, after 2 seconds, i.e. at 5 seconds, the ball should be at its maximum.

h(t = 5) = 18sin(5π/2) + 32 = 18 + 32 = 50, which satisfies the condition.

Hence, option (D) is the right answer

6 0
3 years ago
If one outlier is excluded from a data set, what must happen to the median?
anzhelika [568]
I think that the median decreases, but not so sure.
8 0
3 years ago
FIND X PLZZZZZ I NEED HELP BADLY IDK THE ANSWER OR HOW TO DO IT
svetoff [14.1K]
I suppose we have to find the lengh of the sides.
Area = lenght × width
288 = (3x) × (2x+10)
288 = 3x (2x+10)
288 = 6x + 30x
288 = 36x
288/36 = x
8 = x

therefore the sides are

3x = 3×8 = 24 inches
2x+10 = 2×8 + 10 = 16 +10 = 26 inches
7 0
3 years ago
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