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4 years ago

Find the area of the triangle SHOW WORK

Mathematics

1 answer:

Molodets [167]4 years ago

5 0

Answer:

1236.6 mm

Step-by-step explanation:

formula is 1/2 (b)(h)

1/2(54)(46.8)

since they are all equal sides

the type of triangle is equilateral triangle

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dalvyx [7]

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3 years ago

Simplify the expression:<br> (-2-2i)(-4+6i)

Maurinko [17]

(-2-2i)(-4+6i)
=8-12i+8i-12i^2
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=4(i-1)(3i+2)

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3 years ago

Read 2 more answers

Sofía usted 1/3 of an ounce of cheese to make 1/12 of a pound of pizza dough , how many ounce of cheese are needed to make a pou

Vlad1618 [11]

1/3 • 12= 12/3. 12/3=4. So she needs 4 ounces of cheese for a pound of pizza.

4 0

4 years ago

A circular flower bed is 22 m in diameter and has a circular sidewalk around it that is 4 m wide. Find the area of the sidewalk

antoniya [11.8K]

Answer:

326.6 m^2

to nearest tenth.

Step-by-step explanation:

The radius of the flower bed = 1/2 * 22 = 11m

The radius of the whole area included the sidewalk = 11 + 4 = 15 m

So the area of the sidewalk = area of the whole - area of flower bed

= 3.14 * 15^2 - 3.14* 11^2

= 326.6 m^2.

6 0

3 years ago

A contaminant is leaking into a lake at a rate of R(t) = 1700e^0.06t gal/h. Enzymes that neutralize the contaminant have been ad

olasank [31]

Answer:

16,460 gallons

Step-by-step explanation:

This is a differential equation problem, we have a constant flow of contaminant into the lake, but also we know that only a fraction of that quantity of contaminant remains because of the enzymes. For that reason, the differential equation of contaminant's flow into the lake would be:

$\frac{dQ}{dt} =1700exp(0.06t)*exp(-0.32t)\\\frac{dQ}{dt} =1700exp(-0.26t)\\$

Then, we have to integrate in order to find the equation for Q(t), as the quantity of contaminant in the lake, in function of time.

$\int\limits^0_t {dQ}=\int\limits^0_t {1700exp(-0.26t)dt}\\Q(t)=\frac{1700}{-0.26} exp(-0.26t)+C \\$

Now, we use the given conditions to replace them in the equation, in order to solve for $C$

$t_{0} =0\\Q_{0}=10,000\\Q_{0}=-6538exp(-0.26*0)+C\\C=10,000+6538=16538$

Then, we reorganize the equation and we replace t for 17 hours, in order to determine the quantity of contaminant at that time:

$Q_{t} =-6538exp(-0.26t)+16538\\Q_{17} =-6538exp(-0.26*17)+16538\\Q_{17} =16460 gallons$

3 0

3 years ago