All categories

3 years ago

Help, short on time rn

Mathematics

1 answer:

mafiozo [28]3 years ago

6 0

Two angles are said to be complementary if the some of those two angles is equal to 90°

Given : ∠WXY and ∠YXZ are complementary angles

⇒ ∠WXY + ∠YXZ = 90°

Given : ∠WXY = (2x + 5)° and ∠YXZ = (8x - 5)°

⇒ (2x + 5)° + (8x - 5)° = 90°

⇒ 10x = 90°

⇒ x = 9°

Substituting the value x = 9° in angles ∠WXY and ∠YXZ, we get :

⇒ ∠WXY = 2(9) + 5 = 18 + 5 = 23°

⇒ ∠YXZ = 90° - 23° = 67°

You might be interested in

Two figures are similar, and the scale factor is 2/3

d1i1m1o1n [39]

Answer:

The answer is 8.

Step-by-step explanation:

The scale factor between the figures is 2/3, this means that the ratio of the smaller figure to the larger figure is 2/3:

$\frac{smaller\:figure}{larger\:figure}=\frac{2}{3}$ .

So when a side of the the larger figure is 12 then:

$\frac{smaller\:figure\: side}{12}=\frac{2}{3}$

Therefore

$smaller\:figure\:side=\frac{2}{3}*12=8$ .

Thus the length of the corresponding smaller side is 8.

3 0

3 years ago

Read 2 more answers

Please help me I only have 5 minutes left

Sveta_85 [38]

Answer:

1st one

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Blueberries contain sugars like glucose. What is the source of carbon for the glucose in blueberries?

Kobotan [32]

B. Carbon dioxide gas in air

4 0

3 years ago

Which graph shows the solution to the system of linear inequalities below?

Tasya [4]

Since there isn't a picture of the graphs, I attached one, of my graph, to this answer.

I'm also assuming (y2 2x+1) is actually (y< 2x+1).

Possible solutions are within the purple section of the attached image.

I hope this is helpful to you.

5 0

3 years ago

Which two values of x are roots of the polynomial below? x2-11x + 15

Alex787 [66]

The two values of roots of the polynomial $x^{2}-11 x+15$ are $\frac{11+\sqrt{61}}{2} \text { or } \frac{11-\sqrt{61}}{2}$

Solution:

Given, polynomial expression is $x^{2}-11 x+15$

We have to find the roots of the given expression.

In order to find roots, now let us use quadratic formula.

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Given that $x^{2}-11 x+15$

Here a = 1, b = -11 and c = 15

On substituting the values we get,

$x=\frac{-(-11) \pm \sqrt{(-11)^{2}-4 \times 1 \times 15}}{2 \times 1}$

$\begin{array}{l}{x=\frac{11 \pm \sqrt{121-60}}{2}} \\\\ {x=\frac{11 \pm \sqrt{61}}{2}} \\\\ {x=\frac{11+\sqrt{61}}{2} \text { or } \frac{11-\sqrt{61}}{2}}\end{array}$

Hence, the roots of given polynomial are $\frac{11+\sqrt{61}}{2} \text { or } \frac{11-\sqrt{61}}{2}$

6 0

3 years ago