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3 years ago

How do i know if a system of equation has infinite number of solutions

1 answer:

3 0

A system of equations has an infinite number of solutions when after solving the system, you end up with a true statement.

For example, let’s say you solve a system of equations and your result is 5=5. 5=5 is a true statement, but it doesn’t tell us anything about the x or y value solutions to the system. So because we’re only left with 5=5, there are infinite solutions to the system.

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Can someone please help me with this (it says I have to choose more than 1 answers)

Brrunno [24]

Answer:

B & C

Step-by-step explanation:

Since x is greater than 38, 40 and 120 matches the condition.

Hope u understand.

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7 0

3 years ago

Can someone help please i’ll mark branliest!!

andre [41]

Answer:

a= 32 degrees

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3 years ago

Urgent !!!!! triangle abc in the graph is a right triangle

cluponka [151]

Answer:

The third one

Step-by-step explanation:

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3 years ago

Read 2 more answers

A rectangle has an area of 19.38 cm2. When both the length and width of the rectangle are increased by 1.50 cm, the area of the

Lostsunrise [7]

Answer: $5.7\ cm$

Step-by-step explanation:

Given

Rectangle has an area of $19.38\ cm^2$

Suppose rectangle length and width are $l$ and $w$

If each side is increased by $1.50\ cm$

Area becomes $A_2=35.28\ cm^2$

We can write

$\Rightarrow lw=19.38\quad \ldots(i)\\\\\Rightarrow (l+1.5)(w+1.5)=35.28\\\Rightarrow lw+1.5(l+w)+1.5^2=35.28\\\text{use (i) for}\ lw\\\Rightarrow 19.38+1.5(l+w)=35.28-2.25\\\Rightarrow l+w=9.1\quad \ldots(ii)$

Substitute the value of width from (ii) in equation (i)

$\Rightarrow l(9.1-l)=19.38\\\Rightarrow l^2-9.1l+19.38=0\\\\\Rightarrow l=\dfrac{9.1\pm\sqrt{(-9.1)^2-4(1)(19.38)}}{2\times 1}\\\\\Rightarrow l=\dfrac{9.1\pm\sqrt{5.29}}{2}\\\\\Rightarrow l=\dfrac{9.1\pm2.3}{2}\\\\\Rightarrow l=3.4,\ 5.7$

Width corresponding to these lengths

$w=5.7,\ 3.4$

Therfore, we can write the length of the longer side is $5.7\ cm$

8 0

3 years ago

Find the perimeter of pentagon STUVW WITH VERTICES S(0,0) T(3,-2) U(2,-5) V(-2,-5) W(-3,-2)

andrew11 [14]

Use the distance formula.

$\sqrt{( x_{2} - x_{1} )^2 + (y_{2} - y_{1})^2}$

Points S and W.
$\sqrt{(3)^2 + (2)^2}$

$\sqrt{9+4}$

$\sqrt{13}$

~3.6

Points S and T
$\sqrt{(3 - 0)^2 + (-2 - 0)^2}$

$\sqrt{(3)^2 + (-2)^2}$

$\sqrt{9+4}$

$\sqrt{13}$

~3.6

Points T and U
$\sqrt{(3 - 2)^2 + (-2 + 5)^2}$

$\sqrt{(1)^2 + (3)^2}$

$\sqrt{1+9}$

$\sqrt{10}$

~3.1

Points U and V
$\sqrt{(2+2)^2 + (-5 + 5)^2}$

$\sqrt{(4)^2 + (0)^2}$

$\sqrt{16}$

~4

Points V and W
$\sqrt{(-2+3)^2 + (-5 + 2)^2}$

$\sqrt{(1)^2 + (-3)^2}$

$\sqrt{2+9}$

$\sqrt{11}$

~3.3

Add all these together.

3.3 + 3.1 + 4 + 3.1 + 3.6
≈17

4 0

4 years ago