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s2008m [1.1K]
3 years ago
9

Theodore has a candy jar that is shaped like a triangular prism. The triangular base has a base of 6 cm and a height of 4 cm. If

the jar can hold 60 cm³ of candy, how tall is the jar?
Mathematics
1 answer:
Alex17521 [72]3 years ago
3 0
Use the formula for the volume of a triangular prism:

\sf V=\dfrac{1}{2}ach

Where 'a' is the triangular base base, 'c' is the triangular base height, and 'h' is the height.

Plug in what we know:

\sf 60=\dfrac{1}{2}(6)(4)h

Simplify:

\sf 60=12h

Divide 12 to both sides:

\sf h=\boxed{\sf 5}
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HELPPPPP
OlgaM077 [116]

Answer:

Point on Midline (0,3)

Maximum (9π/2,3)

Minimum (-9π/2,-3)

Step-by-step explanation:

in the given sine function which is in the form of f(x) = a sin(bx+c) +d

a = amplitude

period = frequency = 18π

Therefore b = 2π/18π = 1/9

Y intercept = vertical shift = 3

Horizontal shift = d = 0

Therefore the sine function will be

f(x) = 6 sin(x/9) + 3

Now first point on the midline is (0,3)

Second point is maximum (9π/2,9)

Third point be a minimum value ( -9π/2,-3)

7 0
3 years ago
4. The area of the parallelogram<br>is 81.9 in? What is the height​
ipn [44]

Answer 96in2

Step-by-step explanation:

formula to find area = base x height

substitute the values

= 10 x 9

= 90 cm2

7 0
4 years ago
Find the range of the data in the box plot below.​
harkovskaia [24]

Answer:

im pretty sure its 2

Step-by-step explanation:

8 0
4 years ago
30 POINTS HELPPPPPP
riadik2000 [5.3K]

Answer:

$20.40

Step-by-step explanation:

4 0
3 years ago
Can someone please help me on number 16-ABC
melomori [17]

Answer:

Please check the explanation.

Step-by-step explanation:

Given the inequality

-2x < 10

-6 < -2x

<u>Part a) Is x = 0 a solution to both inequalities</u>

FOR  -2x < 10

substituting x = 0 in -2x < 10

-2x < 10

-3(0) < 10

0 < 10

TRUE!

Thus, x = 0 satisfies the inequality -2x < 10.

∴ x = 0 is the solution to the inequality -2x < 10.

FOR  -6 < -2x

substituting x = 0 in -6 < -2x

-6 < -2x

-6 < -2(0)

-6 < 0

TRUE!

Thus, x = 0 satisfies the inequality -6 < -2x

∴ x = 0 is the solution to the inequality -6 < -2x

Conclusion:

x = 0 is a solution to both inequalites.

<u>Part b) Is x = 4 a solution to both inequalities</u>

FOR  -2x < 10

substituting x = 4 in -2x < 10

-2x < 10

-3(4) < 10

-12 < 10

TRUE!

Thus, x = 4 satisfies the inequality -2x < 10.

∴ x = 4 is the solution to the inequality -2x < 10.

FOR  -6 < -2x

substituting x = 4 in -6 < -2x

-6 < -2x

-6 < -2(4)

-6 < -8

FALSE!

Thus, x = 4 does not satisfiy the inequality -6 < -2x

∴ x = 4 is the NOT a solution to the inequality -6 < -2x.

Conclusion:

x = 4 is NOT a solution to both inequalites.

Part c) Find another value of x that is a solution to both inequalities.

<u>solving -2x < 10</u>

-2x\:

Multiply both sides by -1 (reverses the inequality)

\left(-2x\right)\left(-1\right)>10\left(-1\right)

Simplify

2x>-10

Divide both sides by 2

\frac{2x}{2}>\frac{-10}{2}

x>-5

-2x-5\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-5,\:\infty \:\right)\end{bmatrix}

<u>solving -6 < -2x</u>

-6 < -2x

switch sides

-2x>-6

Multiply both sides by -1 (reverses the inequality)

\left(-2x\right)\left(-1\right)

Simplify

2x

Divide both sides by 2

\frac{2x}{2}

x

-6

Thus, the two intervals:

\left(-\infty \:,\:3\right)

\left(-5,\:\infty \:\right)

The intersection of these two intervals would be the solution to both inequalities.

\left(-\infty \:,\:3\right)  and \left(-5,\:\infty \:\right)

As x = 1 is included in both intervals.

so x = 1 would be another solution common to both inequalities.

<h3>SUBSTITUTING x = 1</h3>

FOR  -2x < 10

substituting x = 1 in -2x < 10

-2x < 10

-3(1) < 10

-3 < 10

TRUE!

Thus, x = 1 satisfies the inequality -2x < 10.

∴ x = 1 is the solution to the inequality -2x < 10.

FOR  -6 < -2x

substituting x = 1 in -6 < -2x

-6 < -2x

-6 < -2(1)

-6 < -2

TRUE!

Thus, x = 1 satisfies the inequality -6 < -2x

∴ x = 1 is the solution to the inequality -6 < -2x.

Conclusion:

x = 1 is a solution common to both inequalites.

7 0
3 years ago
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