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3 years ago

8. Given triangle AXTB with x = 8, t = 11 and b = 14, what is the mZT?

Mathematics

1 answer:

Ber [7]3 years ago

5 0

I need the answer aswell

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PLEASE HELPPP!!!! y = 2/5x + 2

Snezhnost [94]

Answer:

Your answer is y.

Step-by-step explanation:

You start graphing by plotting the 2 and the you go two points up and 5 to the right.

6 0

3 years ago

Help me lol !! will give brainleist

katovenus [111]

Answer:

Step-by-step explanation:

Since, Sum of triangle is equal to 180°

180 = 2x + 50 + 3x - 10

180 = 5x + 40

180 - 40 = 5x

140 = 5x

140/5 = x

28 = x

3 0

2 years ago

Read 2 more answers

Martin says that f(x)=2(4)^x starts at 4 and has a constant ratio of 2. What error did Martin make? Explain.

grigory [225]

Answer:

A general exponential equation is written as:

f(x) = A*(r)^x

Where:

A is the initial value, this is the value that the function takes when x = 0.

r is the rate of growth

x is the variable.

In this case, the function is:

f(x) = 2*(4)^x

The initial value is:

f(0) = 2*(4)^0 = 2*1 = 2

The initial value is 2.

And the rate of growth is 4.

So Martin seems to mixed these two values (he said initial value = 4, and rate = 2, which is the opposite of what we found)

8 0

2 years ago

patty is building a rope ladder for a tree house. she needs two 5-foot pieces of rope for th esides fo the ladder. she needs 7 p

olga nikolaevna [1]

8 because 8comes after 5 and 7

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2 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