All categories

4 years ago

Which relationship is always correct for the angles p, q, and r of triangle ABC?

Mathematics

1 answer:

irakobra [83]4 years ago

8 0

The correct answer is "D".

q+r=p

Because the missing angle would equal 180-q-r, it would be the same as p. The one angle (the unmarked) is supplementary to the missing angle. (180 degrees)

I hope this helps!
~kaikers

You might be interested in

Helpppppppp easyyyyyyy

Margaret [11]

awnser is 20 and awnser a

3 0

4 years ago

Is the correctx 25<28

trapecia [35]

Answer:

Yes

Step-by-step explanation:

25 is less than 28

28 is greater than 25

4 0

3 years ago

Read 2 more answers

Find the slope of the line that passes through (6 , 2) and ( 1 ,10).

DochEvi [55]

Answer: Slope: -1.6

Use the slope formula

substitute point values in the formula

(10 -2)

- like this: m = ----------

(1 - 6)

Simplify each other side of the equation

- simplified: 8/-5

m: - 1.6

7 0

3 years ago

A relation is (1 point) Select one: a. the output (y) values of the relation b. the input (x) values of the relation c. a set of

ArbitrLikvidat [17]

Answer:

a. the output (y) values of the relation

Step-by-step explanation:

The range of a function is the set of all possible values that a dependent variable would take.

Therefore, it is the output values of the relation.

3 0

3 years ago

Read 2 more answers

F(x)=x^2-3x+18 how many distinct real number zeros does f have?

vampirchik [111]

Answer:

None of the distinct real number zeros the function have.

Step-by-step explanation:

Considering the function

$f(x)=x^2-3x+18$

The zeroes of a function are those values that touches the x-axis. In order to find those values we must

Set $f(x) = 0$ in order to find those values

$0=x^2-3x+18$

$\mathrm{Switch\:sides}$

$x^2-3x+18=0$

$\mathrm{Solve\:with\:the\:quadratic\:formula}$

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=1,\:b=-3,\:c=18:\quad x_{1,\:2}=\frac{-\left(-3\right)\pm \sqrt{\left(-3\right)^2-4\cdot \:1\cdot \:18}}{2\cdot \:1}$

$x=\frac{-\left(-3\right)+\sqrt{\left(-3\right)^2-4\cdot \:1\cdot \:18}}{2\cdot \:1}$

$x=\frac{3+\sqrt{\left(-3\right)^2-4\cdot \:1\cdot \:18}}{2\cdot \:1}$

$3+\sqrt{\left(-3\right)^2-4\cdot \:1\cdot \:18}=3+\sqrt{63}i$

$x=\frac{3+\sqrt{63}i}{2\cdot \:1}$

$x=\frac{3+3\sqrt{7}i}{2}$

$x=\frac{3}{2}+\frac{3\sqrt{7}}{2}i$

Similarly,

$x=\frac{-\left(-3\right)-\sqrt{\left(-3\right)^2-4\cdot \:1\cdot \:18}}{2\cdot \:1}:\quad \frac{3}{2}-i\frac{3\sqrt{7}}{2}$

$\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$

$x=\frac{3}{2}+i\frac{3\sqrt{7}}{2},\:x=\frac{3}{2}-i\frac{3\sqrt{7}}{2}$

BUT, NONE OF THEM ARE REAL NUMBER ZEROS.

Therefore, none of the distinct real number zeros the function have.

5 0

4 years ago