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3 years ago

Is 10(x-2)=5(2x-4) infinitely many or no solution

Mathematics

2 answers:

Andrei [34K]3 years ago

7 0

10x-20=10x-20, make your decision on the answer from this which they are all real numbers. Hope this will help you!

anzhelika [568]3 years ago

4 0

10(x-2)=5(2x-4)
10x - 20 = 10x - 20

Here, As L.H.S = R.H.S,

System of Equation has Infinitely many Solution

Hope this helps!

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Express as a difference: a+b, a+5

Tcecarenko [31]

Answer:

a - (-b), a - (-5)

Step-by-step explanation:

A double negative negative is a positive, so a - (-b) = a + b, and a - (-5) is a + 5. However, a double negative is still a difference, so this answer works.

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3 years ago

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yarga [219]

1st option.

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3 years ago

From the top of a 100-ft building a man

romanna [79]

Answer:

219m

Step-by-step explanation:

Since the man observes the car with angle 15 before observing in 33 degrees.

For the first observation

The angle observation gives an angle if 33 degrees with the horizontal.

It gives a triangle which I'll attach to the que,

from the first triangle

Tan 33 = 100/y

Y= 100/ tan 33

Y = 153.99m.

This is the distance from the building to the distance where it was secondly observed( 33).

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153.99 + x = 100/ tan 15

153.99 + x = 373.21

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3 years ago

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3 years ago

Read 2 more answers

Write a definite integral that represents the area of the region. (Do not evaluate the integral.) y1 = x2 + 2x + 3 y2 = 2x + 12F

Svet_ta [14]

Answer:

$A = \int\limits^3__-3}{9}-{x^{2}} \, dx = 36$

Step-by-step explanation:

The equations are:

$y = x^{2} + 2x + 3$

$y = 2x + 12$

The two graphs intersect when:

$x^{2} + 2x + 3 = 2x + 12$

$x^{2} = 0$

$x_{1} = 3\\x_{2} = -3$

To find the area under the curve for the first equation:

$A_{1} = \int\limits^3__-3}{x^{2} + 2x + 3} \, dx$

To find the area under the curve for the second equation:

$A_{2} = \int\limits^3__-3}{2x + 12} \, dx$

To find the total area:

$A = A_{2} -A_{1} = \int\limits^3__-3}{2x + 12} \, dx -\int\limits^3__-3}{x^{2} + 2x + 3} \, dx$

Simplifying the equation:

$A = \int\limits^3__-3}{2x + 12}-({x^{2} + 2x + 3}) \, dx = \int\limits^3__-3}{9}-{x^{2}} \, dx$

Note: The reason the area is equal to the area two minus area one is that the line, area 2, is above the region of interest (see image).

3 0

2 years ago