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3 years ago

Find the surface area of the equilateral triangular pyramid.

Mathematics

1 answer:

Alenkinab [10]3 years ago

8 0

We know that
[surface area of the triangular pyramid]=area of the base+3*[area of lateral triangles]
area of the base=12*8/2-----> 48 cm²

area of one lateral triangle=12*10/2-----> 60 cm²
[surface area of the triangular pyramid]=48+3*[60]-----> 228 cm²

the answer is the option
B) 228 cm2

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Justify each step of this inequality by stating the property that was used to get to each step.

Allushta [10]

Answer:

Step-by-step explanation:

step 1 simplified like terms -11k +3k =8k

step 2 incorrect should be -4k -20 on the left side distributive property

step 3 added 8k to each side

step 4 added 20 to each side

step 5 divided each side by 4

6 0

3 years ago

What is the y-value when x equals 28? y=310-25(x)

Fofino [41]

Answer:

-390 PLEASE GIVE BRAINLIEST

Step-by-step explanation:

y = 310 - 25(28)

y = 310 - 700

y = -390

5 0

3 years ago

A club raised 175% of its goal for a charity. The club raises $875. What was the goal?

charle [14.2K]

Answer: The goal was $500

======================================================

Work Shown:

x% = x/100

175% = 175/100 = 1.75

Let g be the goal, which is the amount of money the club wanted to raise

175% of goal = 175% of g = 1.75g

The expression 1.75g represents how much money was actually raised, which was $875. Set the two expressions equal to each other. Solve for g

1.75g = 875

g = 875/1.75 ....... divide both sides by 1.75

g = 500

The club's goal was to raise $500

Note how 75% of $500 is 0.75*500 = 375

When they raised 175% of the goal, this means they went 75% overboard and added on 375 additional dollars (on top of the 500 they wanted). So they got to 500+375 = 875 which lines up with the instructions. This helps verify the answer.

Or we can see that 1.75*g = 1.75*500 = 875 which helps confirm the answer as well.

3 0

3 years ago

What is the value of y?

alexira [117]

9514 1404 393

Answer:

y = 8√3

Step-by-step explanation:

In the 30°-60°-90° "special" triangle, the ratio of sides lengths is ...

1 : √3 : 2 = 8 : y : hypotenuse

In order for these ratios to be equal, we must have ...

y = 8√3

If you want to solve this using your trig skills, you recognize that ...

Tan = Opposite/Adjacent

tan(30°) = 8/y

y = 8/tan(30°) . . . . where tan(30°) = 1/√3

y = 8√3

4 0

3 years ago

For the given term, find the binomial raised to the power, whose expansion it came from: 15(5)^2 (-1/2 x) ^4

Elina [12.6K]

Answer:

C. $(5-\frac{1}{2})^6$

Step-by-step explanation:

Given

$15(5)^2(-\frac{1}{2})^4$

Required

Determine which binomial expansion it came from

The first step is to add the powers of he expression in brackets;

$Sum = 2 + 4$

$Sum = 6$

Each term of a binomial expansion are always of the form:

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

Where n = the sum above

$n = 6$

Compare $15(5)^2(-\frac{1}{2})^4$ to the above general form of binomial expansion

$(a+b)^n = ......+15(5)^2(-\frac{1}{2})^4+.......$

Substitute 6 for n

$(a+b)^6 = ......+15(5)^2(-\frac{1}{2})^4+.......$

[Next is to solve for a and b]

From the above expression, the power of (5) is 2

Express 2 as 6 - 4

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

By direct comparison of

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

and

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

We have;

$^nC_ra^{n-r}b^r= 15(5)^{6-4}(-\frac{1}{2})^4$

Further comparison gives

$^nC_r = 15$

$a^{n-r} =(5)^{6-4}$

$b^r= (-\frac{1}{2})^4$

[Solving for a]

By direct comparison of $a^{n-r} =(5)^{6-4}$

$a = 5$

$n = 6$

$r = 4$

[Solving for b]

By direct comparison of $b^r= (-\frac{1}{2})^4$

$r = 4$

$b = \frac{-1}{2}$

Substitute values for a, b, n and r in

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

$(5+\frac{-1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

Solve for $^6C_4$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{(6-4)!4!)}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{2!!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5*4!}{2*1*!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5}{2*1}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{30}{2}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^2(\frac{-1}{2})^4+.......$

Check the list of options for the expression on the left hand side

The correct answer is $(5-\frac{1}{2})^6$

3 0

3 years ago