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3 years ago

0.2x + 0.5y = 4 -0.1x + 0.3y = -2

Mathematics

2 answers:

crimeas [40]3 years ago

5 0

Answer:

A. (20, 0)

Step-by-step explanation:

riadik2000 [5.3K]3 years ago

3 0

Your solving for variable x
do move all the terms containing x to the left and all other terms to the right

so we going to add -0.5y to each side of the equation
0.2x + 0.5y + -0.5y = 4 + -0.5y

them combine like terms 0.5y + -0.5y =0.0
0.2x +0.0 =4 + -0.5y
0.2x = 4 +-0.5y

Divide each side by 0.2
x= 20 +-2.5y

x=20+ -2.5y

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Can someone please help me answer these

Anni [7]

Step-by-step explanation:

5: odd

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3 years ago

A²,b²,c² are consecutive perfect squares.How many natura numbers are lying between a² and c², if a>0

pogonyaev

Answer:

The quantity of natural numbers between $a^{2}$ and $c^{2}$ is $2\cdot (a + b) + 1$ .

Step-by-step explanation:

If $a^{2}$ , $b^{2}$ and $c^{2}$ are consecutive perfect squares, then both $a$ , $b$ and $c$ are natural numbers and we have the following quantities of natural numbers:

Between $b^{2}$ and $c^{2}$ :

$c^{2} = (b+1)^{2}$

$c^{2} = b^{2}+2\cdot b + 1$

$c^{2}-b^{2} = 2\cdot b + 1$

And the quantity of natural numbers between $b^{2}$ and $c^{2}$ is:

$c^{2}-b^{2}-1 = 2\cdot b$

Between $a^{2}$ and $b^{2}$ :

$b^{2} = (a + 1)^{2}$

$b^{2} = a^{2} +2\cdot a + 1$

$b^{2}-a^{2} = 2\cdot a + 1$

And the quantity of natural numbers between $a^{2}$ and $b^{2}$ is:

$b^{2}-a^{2}-1 = 2\cdot a$

And the quantity of natural numbers between $a^{2}$ and $c^{2}$ is:

$Diff = 2\cdot a + 2\cdot b + 1$

Please observe that the component +1 represents the natural number $b^{2}$

4 0

3 years ago

Which of the following equations are true for the number 5? (Select all that apply)

postnew [5]

Answer:

A. True

B. True

C. False

D. False

6 0

3 years ago

Kristen lives directly east of the park. The football field is directly south of the park. The library sits on the line formed b

klemol [59]

Imagine right triangle PHF, where P - park, H - home and F - football field, then PH, PF are legs and HF is hypotenuse . Denote point L to be library. You know that point L lies on the segment FH and FL=8, LH=2. Also you know that PL is an altitude to the hypotenuse.

Use the property of altitude drawn from the vertex of right angle to the hypotenuse (the length of the altitude is geometrical mean between legs' projections onto hypotenuse):

$PL=\sqrt{HL\cdot LF},\\ PL=\sqrt{2\cdot 8}=\sqrt{16} =4$ mi.

This means that the distance between park and libriry is 4 miles.

Consider right triangle PLF ( angle L is right angle and PF - hypotenuse). By the Pythagorean theorem,

$PF^2=PL^2+LF^2,\\ PF^2=4^2+8^2,\\ PF^2=16+64=80,\\ PF=\sqrt{80} =4\sqrt{5}$ mi. The distance between park and football field is $4\sqrt{5}$ miles.

Answer: the distance between park and libriry is 4 miles and the distance between park and football field is $4\sqrt{5}$ miles.

6 0

3 years ago

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s344n2d4d5 [400]

Answer:

Step-by-step explanation:

# angle ABC is congruent to angle DEF which’s given

# angle GHI is congruent to DEF which’s given

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6 0

3 years ago