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Maurinko [17]
3 years ago
6

-4+8(7+×) is equivalent to

Mathematics
1 answer:
Helen [10]3 years ago
7 0
-4+56+8x= 8x+52
8x+1w is the answer
You might be interested in
Please solve with working thankyou
Lisa [10]

Answer:

m∠y=72°

Step-by-step explanation:

<u><em>The question is</em></u>

Find the measure of angle y

step 1

Find the measure of angle x

see the attached figure to better understand the problem

we know that

m∠x=67° -----> by corresponding angles

step 2

Find the measure of angle y

we know that

(m∠x+m∠y)+42°=180° -----> because form a straight line

substitute the measure of angle x

67°+m∠y+42°=180

m∠y+108°=180

m∠y=180°-108°

m∠y=72°

4 0
3 years ago
Prove that the diagonals of a parallelogram bisect each other.<br> The midpoint of AC is
iren2701 [21]

Answer:

Theorem: The diagonals of a parallelogram bisect each other. Proof: Given ABCD, let the diagonals AC and BD intersect at E, we must prove that AE ∼ = CE and BE ∼ = DE. The converse is also true: If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram.

Step-by-step explanation:

3 0
3 years ago
In engineering, equilateral triangles can support the most weight and so are commonly found in the design of bridges and buildin
Dafna1 [17]

Answer:

60°

Step-by-step explanation:

Am equilateral triangle is a triangle that has all its sides and angles equal. Note that an equilateral triangle has 3 sides and angles.

Let the angles be a, b and c

The sum of angle in a triangle is 180°, hence;

a+b+c = 180°

Since a = b = c

The equation becomes;

a+a+a = 180°

3a = 180

a = 180/3

a = 60°

Hence the measure of the angles of an equilateral triangle is 60° for the three angles.

4 0
3 years ago
find the component equation of the plane which is normal to the vector -2i+5j+k and which contains the point (-10;7;5).​
maksim [4K]

Given:

A plane is normal to the vector = -2i+5j+k

It contains the point (-10,7,5).​

To find:

The component equation of the plane.

Solution:

The equation of plane is

a(x-x_0)+b(y-y_0)+c(z-z_0)=0

Where, (x_0,y_0,z_0) is the point on the plane and \left< a,b,c\right> is normal vector.

Normal vector is -2i+5j+k and plane passes through (-10,7,5). So, the equation of the plane is

-2(x-(-10))+5(y-7)+1(z-5)=0

-2(x+10)+5y-35+z-5=0

-2x-20+5y-35+z-5=0

-2x+5y+z-60=0

-2x+5y+z=60

Therefore, the equation of the plane is -2x+5y+z=60.

4 0
3 years ago
50 points please help me
skelet666 [1.2K]

Answer:

0

Step-by-step explanation:

The slope of the line is 0 as it is parallel to x-axis.

Slope = (y2 - y1)/(x2 - x1)

= (2-2)/(4-(-2))

= 0

8 0
3 years ago
Read 2 more answers
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