All categories

3 years ago

I need help with this and how you got it please! 10 points

Mathematics

1 answer:

Elis [28]3 years ago

4 0

Answer:

B, x=40 is the right answer

Step-by-step explanation:

DAC is one half of the right angle and CAB is another half. So, you know that since a right angle is 90 degrees, DAC + CAB must equal 90.

They have given you equations for both angles, so now its just a matter of rearranging and solving for x.

DAC = x

CAB = 2x - 30

DAC + CAB = 90

Using the equations above, we can come up with this final equation:

x + 2x - 30 = 90

If we add the x together:

3x - 30 = 90

If we move the 30 to the other side by adding it to 90

3x = 90 + 30

3x = 120

If we divide 120 by 3:

x= 40

Therefore, the answer is B

Hope this helped : )

You might be interested in

WILL GIVE BRANLIEST!!! Pls help! Determine the coordinates of the point on the straight line y=3x+1 that is equidistant from the

iren [92.7K]

Let , coordinate of points are P( h,k ).

Also , k = 3h + 1

Distance of P from origin :

$d=\sqrt{h^2+k^2}$

Distance of P from ( -3, 4 ) :

$d=\sqrt{(h+3)^2+(k-4)^2}$

Now , these distance are equal :

$h^2+(3h+1)^2=(h+3)^2+(3h+1-4)^2\\\\h^2+(3h+1)^2=(h+3)^2+(3h-3)^2$

Solving above equation , we get :

$P=(\dfrac{16}{21},\dfrac{23}{7})$

Hence , this is the required solution.

6 0

3 years ago

Will mark brainliest! The triangles shown below must be congruent, true or false

olchik [2.2K]

True because they have the same size angles

8 0

3 years ago

Read 2 more answers

Pls, Help !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

professor190 [17]

I can help you if you still!!!

4 0

2 years ago

In a recent

qwelly [4]

Answer:collecting data

Step-by-step explanation:

7 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