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LUCKY_DIMON [66]

2 years ago

5

PLEASE HELP ASAP. GIVE ANSWER AND BRIEF EXPLANATION.

1 answer:

scoray [572]2 years ago

4 0

A. hope this helps):)

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Which value, when placed in the box, would result in a system of equations with infinitly many solutions? Y=2x - 5 2y-4x=

Nezavi [6.7K]

Given:

The system of equations is

$y=2x-5$

$2y-4x=?$

To find:

The missing value for which the given system of equations have infinitely many solutions.

Solution:

Let the missing value be k.

We have,

$y=2x-5$

$2y-4x=k$

Taking all the terms on the left side, the given equations can be rewritten as

$-2x+y+5=0$

$-4x+2y-k=0$

The system of equations $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ have infinitely many solutions if

$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$

We have,

$a_1=-2,b_1=1,c_1=5$

$a_2=-4,b_2=2,c_2=-k$

Now,

$\dfrac{-2}{-4}=\dfrac{1}{2}=\dfrac{5}{-k}$

$\dfrac{1}{2}=\dfrac{1}{2}=\dfrac{5}{-k}$

$\dfrac{1}{2}=\dfrac{5}{-k}$

On cross multiplication, we get

$-k=10$

$k=-10$

Therefore, the missing value is -10.

6 0

2 years ago

Which of the following is a list of unlike terms, or terms that cannot be combined?

Vilka [71]

Answer:

C

Step-by-step explanation:

C is the only choice where all given options have different or no coefficients.

3 0

2 years ago

Demerol 45mg and atropine 400mcg/ml give how much per ml to total volume to inject demerol contains 50mg/ml, atropine contains 4

OverLord2011 [107]

Answer:

Volume injected = 1.65 ml

Step-by-step explanation:

mass of Demerol =45 mg.

density of pre-filled in syringe = 50 mg/ml

Volume = 45/50 = 0.9 ml

For atropine mass = 0.3 mg

density= 400 mcg/ml [ Note : 1 mcg = 0.001 mg]

volume filled = 0.3/400(0.001) = 0.75 ml

So, the total volume filled = 0.75+0.9 = 1.65 ml

6 0

2 years ago

Find absolute max and min on a given interval calculator

olga nikolaevna [1]

Min {sin(x) | 0 is les than or equal to x less than or greater than or equal to 2 pie = -1 at x = 3 pie over 2

7 0

3 years ago

Una piscina rectangular de 15 metros de largo por 9 metros de ancho està rodeada por un

stira [4]

Answer: 2 meters.

Step-by-step explanation:

Let w = width of the cement path.

Dimensions of pool : Length = 15 meters , width = 9 meters

Area of pool = length x width = 15 x 9 = 135 square meters

Along width cement path, the length of region = $2w+15$

width = $2w+9$

Area of road with pool = $(2w+15) (2w+9)$

$= 4w^2+30w+18w+135\\\\=4w^2+48w+135$

Area of road = (Area of road with pool ) -(area of pool)

$\Rightarrow\ 112 =4w^2+48w+135- 135\\\\\Rightarrow\ 112= 4w^2+48w\\\\\Rightarrow\ 4 w^2+48w-112=0\\\\\Rightarrow\ w^2+12w-28=0\ \ \ [\text{Divide both sides by 4}]\\\\\Rightarrow\ w^2+14w-2w-28=0\\\\\Rightarrow\ w(w+14)-2(w+14)=0\\\\\Rightarrow\ (w+14)(w-2)=0\\\\\Rightarrow\ w=-14\ or \ w=2$

width cannot be negative, so w=2 meters

Hence, the width of the road = 2 meters.

8 0

3 years ago