All categories

3 years ago

Question 1 Which of the following statements is NOT true?

Mathematics

2 answers:

irakobra [83]3 years ago

6 0

Answer:

The slope of AB is different than the slope of BC

Step-by-step explanation:

My big brain

Olin [163]3 years ago

4 0

Answer:

the bottom one

Step-by-step explanation:

They are both the same ab and bc but im not entirely sure

You might be interested in

What is the unit for the volume of a plastic block

Daniel [21]

The unit is a kilometer.

3 0

3 years ago

The coordinates of A, B, and C in the diagram are A(p,4), B(6,1), and C(9,q). Which equation correctly relates p and q?

Temka [501]

ANSWER

$p + q= 7$

EXPLANATION

We determine the slope of each line using the slope formula;

$m = \frac{y_2-y_1}{x_2-x_1}$

The slope of BC is

$= \frac{ q - 1}{9 - 6}$

$\frac{ q - 1}{3}$

The slope of AB is

$= \frac{1 - 4}{6 - p}$

$= - \frac{3}{6 - p}$

The two lines are perpendicular so the product of their slopes is -1.

$- \frac{3}{6 - p} \times \frac{ q - 1}{3} = - 1$

This implies that,

$\frac{q - 1}{6 - p} = 1$

$q - 1=6 - p$

$q + p = 6 + 1$

$p + q= 7$

4 0

3 years ago

Read 2 more answers

Please help me I don’t understand this

sladkih [1.3K]

Answer :

67.5 degrees

Let the measure of two equal angles of the triangles is x.

Thus, the angles of the triangles are x, x and 45 degrees.

The sum of the measures of the angles of all triangles is 180 degrees.

Hence, the equation is

x + x + 45 = 180

Solve the equation for x

2x + 45 = 180

2x = 180 - 45

2x = 135

x = 67.5

Therefore, measure of other two angles are 67.5 degrees.

8 0

3 years ago

HELP MARKING BRAINLIEST EXTRA POINTS !!

Scrat [10]

Answer:

The quadrilateral is a rectangle
$Perimeter=8\sqrt{2} \,\,Unit$
$Area=6\,\,sq.unit$

Step-by-step explanation:

Given,

$A=(0,0)\\B=(1,1)\\C=(4,-2)\\D=(3,-3)$

Now,

$Slope\,\,for\,\,AB=\frac{1-0}{1-0} =1\\\\Slope\,\,for\,\,CD=\frac{-3+2}{3-4} =1\\\\Slope\,\,for\,\,AD=\frac{-3-0}{3-0} =-1\\\\Slope\,\,for\,\,BC=\frac{-2-1}{4-1} =-1\\\\So,AB||CD\,\,and\,\,AD||BC$

$AB=\sqrt{(1-0)^2+(1-0)^2}=\sqrt{2}\\BC=\sqrt{(4-1)^2+(-2-1)^2}=3\sqrt{2}\\CD=\sqrt{(3-4)^2+(-3+2)^2}=\sqrt{2}\\AD=\sqrt{(3-0)^2+(-3-0)^2}=3\sqrt{2}\\\\So,\,\,AB=CD,AD=BC\\The\,\,quadrilateral\,\,is\,\,a\,\,rectangle$