All categories

2 years ago

Please help me it will be greatly appreciated. will give brainliest. question is in attachment.

Mathematics

1 answer:

Oksi-84 [34.3K]2 years ago

6 0

A) 180-125 =55
B) both bottom angles are 55 and the triangle's angles should add up to 180, so 55 + 55 + x = 180. X = 70
C) CAE and CBD are equal so the bottom angle (that was calculated in A) will be the base acute angle —>55

You might be interested in

When this car moves forward by 180 cm, each wheel does one full turn.

Andre45 [30]

D is the diameter of the wheel
the circle is π.d = 180
so d = 180/π = 57 cm

3 0

3 years ago

Guysssss I'm struggling in mathh :( I need help

Sedaia [141]

Answer:

The first two is x=60

Step-by-step explanation:

1.5/90 = 1/x

1.5x = 90 x 1

1.5x/1.5 = 90/1.5

x = 60

5 0

3 years ago

Read 2 more answers

Nata [24]

Answer:

400,000km is represented as 8cm on the map!!!

Step-by-step explanation:

1cm=50000km

8cm=x

Criss cross

1cmx=8cm×50000km

divide both side with 1cm

x=8×50000km

x=400,000km

5 0

3 years ago

Divide the rational expressions and express in simplest form. When typing your answer for the numerator and denominator be sure

Veseljchak [2.6K]

Dividing by a fraction is equivalent to multiply by its reciprocal, then:

$\begin{gathered} \frac{3y^2-7y-6}{2y^2-3y-9}\div\frac{y^2+y-2}{2y^2+y-3^{}}= \\ =\frac{3y^2-7y-6}{2y^2-3y-9}\cdot\frac{2y^2+y-3}{y^2+y-2}= \\ =\frac{(3y^2-7y-6)(2y^2+y-3)}{(2y^2-3y-9)(y^2+y-2)} \end{gathered}$

Now, we need to express the quadratic polynomials using their roots, as follows:

$ay^2+by+c=a(y-y_1)(y-y_2)$

where y1 and y2 are the roots.

Applying the quadratic formula to the first polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{7\pm\sqrt[]{(-7)^2-4\cdot3\cdot(-6)}}{2\cdot3} \\ y_{1,2}=\frac{7\pm\sqrt[]{121}}{6} \\ y_1=\frac{7+11}{6}=3 \\ y_2=\frac{7-11}{6}=-\frac{2}{3} \end{gathered}$

Applying the quadratic formula to the second polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot2\cdot(-3)}}{2\cdot2} \\ y_{1,2}=\frac{-1\pm\sqrt[]{25}}{4} \\ y_1=\frac{-1+5}{4}=1 \\ y_2=\frac{-1-5}{4}=-\frac{3}{2} \end{gathered}$

Applying the quadratic formula to the third polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{3\pm\sqrt[]{(-3)^2-4\cdot2\cdot(-9)}}{2\cdot2} \\ y_{1,2}=\frac{3\pm\sqrt[]{81}}{4} \\ y_1=\frac{3+9}{4}=3 \\ y_2=\frac{3-9}{4}=-\frac{3}{2} \end{gathered}$

Applying the quadratic formula to the fourth polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot1\cdot(-2)}}{2\cdot1} \\ y_{1,2}=\frac{-1\pm\sqrt[]{9}}{2} \\ y_1=\frac{-1+3}{2}=1 \\ y_2=\frac{-1-3}{2}=-2 \end{gathered}$

Substituting into the rational expression and simplifying:

$\begin{gathered} \frac{3(y-3)(y+\frac{2}{3})2(y-1)(y+\frac{3}{2})}{2(y-3)(y+\frac{3}{2})(y-1)(y+2)}= \\ =\frac{3(y+\frac{2}{3})}{2(y+2)}= \\ =\frac{3y+2}{2y+4} \end{gathered}$

8 0

1 year ago

Which statement is true? All squares are rectangles. All quadrilaterals are rectangles. All parallelograms are rectangles. All r

azamat

Answer:

all squares are rectangles is true

Step-by-step explanation:

4 0

3 years ago