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evablogger [386]
3 years ago
10

in parallelogram abcd angle b is a right angle determine whether the parallelogram is a rectangle if so by what property​

Mathematics
1 answer:
Digiron [165]3 years ago
8 0

Answer:

If a parallelogram has at least one 90 degree angle then its opposite angle is equal and also = 90 degrees.  

The 2 remaining opposite angles add up to 180 degrees and since they must be equal, they must be 90 degrees each.

So, we have a quadrilateral with four 90 degree angles and is therefore a rectangle OR it could possibly be a square.

Step-by-step explanation:

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5x-4y=-10 <br> -4x+5y=8<br> (Khan academy)
Yakvenalex [24]

Answer:

X = 10

Y = 48/5

Step-by-step explanation:

Your welcome.

3 0
3 years ago
The domain of both f(x) = x - 6 and g(x) = x + 6 is all real numbers. What is the domain of h(x) = ?
dolphi86 [110]
The denominator of a fraction cannot be zero.Therefore in h(x) = f(x)/g(x), g(x) cannot equal zero, x + 6 cannot equal zero, x cannot equal -6. The domain of h(x) is ℜ - {-6}.

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
4 0
4 years ago
HELP! HELP!
alexgriva [62]

Answer:

The answer should be C. By combining like terms you can find the answer. 1.75a+2.75a = 4.5a

2.25b + 1.75b = 4b

2.25c + 1.25c = 3.5c

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Urgent! <br><br> 1 1/6 - 3/6 =?
Mnenie [13.5K]

Answer:

1 4/6

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
1. Write the standard form of the line that passes through the given points. (7, -3) and (4, -8)
Sveta_85 [38]

Answer:

1. -5x+3y+44=0

2. 2x+y-2=0

3. 2x+y-4=0

Step-by-step explanation:

Standard form of a line is Ax+By+C=0.

If a line passing through two points then the equation of line is

y-y_1=m(x-x_1)

where, m is slope, i.e.,m=\dfrac{y_2-y_1}{x_2-x_1}.

1.

The line passes through the points (7,-3) and (4,-8). So, the equation of line is

y-(-3)=\dfrac{-8-(-3)}{4-7}(x-7)

y+3=\dfrac{-5}{-3}(x-7)

y+3=\dfrac{5}{3}(x-7)

3(y+3)=5(x-7)

3y+9=5x-35

-5x+3y+9+35=0

-5x+3y+44=0

Therefore, the required equation is -5x+3y+44=0.

2.

We need to find the equation of the line, in standard form, that has a y-intercept of 2 and is parallel to 2 x + y =-5.

Slope of the line : m=\dfrac{-\text{Coefficient of x}}{\text{Coefficient of y}}=\dfrac{-2}{1}=-2

Slope of parallel lines are equal. So, the slope of required line is -2 and it passes through the point (0,2).

Equation of line is

y-2=-2(x-0)

y-2=-2x

2x+y-2=0

Therefore, the required equation is 2x+y-2=0.

3.

We need to find the equation of the line, in standard form, that has an x-intercept of 2 and is parallel to 2x + y =-5.

From part 2, the slope of this line is -2. So, slope of required line is -2 and it passes through the point (2,0).

Equation of line is

y-0=-2(x-2)

y=-2x+4

2x+y-4=0

Therefore, the required equation is 2x+y-4=0.

5 0
3 years ago
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