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3 years ago

Solving a system of linear equations algebraically:

Mathematics

1 answer:

guajiro [1.7K]3 years ago

8 0

It was so easy xD

its x=1

y=12×1+7

y=-6×1+25

=19

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ad-work [718]

Answer:

see explanation

Step-by-step explanation:

a and b lie on the straight line CA, hence

and b sum to 180°, that is supplementary angles

8 0

3 years ago

Read 2 more answers

Help pls I really need it

Talja [164]

Answer:

Infinite number of solutions

Step-by-step explanation:

when you solve for V, you notice that the term in "v" goes away, and you end up with a true statement:

2 = 2

This is a true statement no matter what values the variable V has, so it is true for all possible (infinite) values of "v".

8 0

3 years ago

A rectangular sign has length of 3 and 1/2 ft and a width of 1 and 1/8 ft. Will the area of the sign be greater or less than 3 a

zavuch27 [327]

Answer:

$Area = 3\dfrac{15}{16}\ sq\ ft$

Area is greater than $3\frac{1}{2}$ sq ft.

Step-by-step explanation:

Length of rectangular sign board = $3\frac{1}{2}$ ft

Width of rectangular sign board = $1\frac{1}{8}$ ft

Area of a rectangle is given by the formula:

$A = Length \times Width$

To find the area, we are going to multiply $3\frac{1}{2}$ with $1\frac{1}{8}$ .

$1\frac{1}{8}$ is greater than 1.

Therefore the multiplication will be greater than $3\frac{1}{2}$ .

Hence, we can say that area will be greater than $3\frac{1}{2}$ sq ft.

Area of the sign:

$A = 3\dfrac{1}{2}\times 1\dfrac{1}8\\\Rightarrow A = \dfrac{7}{2}\times \frac{9}8\\\Rightarrow A = \dfrac{63}{16}\\\Rightarrow A = 3\dfrac{15}{16}\ sq\ ft$

8 0

2 years ago

Show both decimal places (5.06).

lilavasa [31]

a. the cost of a call for six minutes is 0.70.

b.the cost of a 14-minute call is 2.62.

c.the cost of a 9 ½2-minute call is 1.66.

given that

rate schedule for an m-minute call from any of its pay phones 0.70.

c(m) = 0.70 when m<=6

c(m) = 0.70+ 0.24(m-6) when m>6 & m is an integer

c(m) = 0.70+0.24((m-6)+1) when m>6 & m is not an integer.

a. to find the cost of a call for six minutes.

already given that c(m) =0.70 when m=6.

Therefore, the cost of a call for six minutes is 0.70.

b. to find the cost of a 14-minute call.

according to given information

$c(14) = 0.70 + 0.24(m - 6) \\ c(14) = 0.70 + 0.24(14 \: - 6) \\c(14) = 0.70 + (0.24 \times 14) - (0.24 \times 6) \\c(14) = 0.70 + (3.36 - 1.44) \\ c(14) = 0.70 + 1.92 \\ c(14) = 2.62$

Therefore,the cost of a 14-minute call is 2.62.

c.to find the cost of a 9 ½2-minute call

according to given information

$c(9 \frac{1}{2} 2) = 0.70 + 0.24((9 \frac{1}{2} 2 - 6) + 1) \\ c(9 \frac{1}{2} 2) = 0.70 + 0.24((9 - 6) + 1) \\ c(9 \frac{1}{2} 2) = 0.70 + 0.24(3+ 1) \\ c(9 \frac{1}{2} 2) = 0.70 + 0.24(4) \\ c(9 \frac{1}{2} 2) = 0.70 + 0.96 \\ c(9 \frac{1}{2} 2) = 1.66$

Therefore,the cost of a 9 ½2-minute call is 1.66.

Learn more about problems on cost of call, refer:

brainly.com/question/16584226

#SPJ9

4 0

1 year ago

Find the positive value of x in the geometric sequence 7x-2,4x+4,3x

Goryan [66]

Hello,

Let's assume k the ratio.

$
 \left \{ {{3x=k*(4x+4)} \atop {4x+4=k*(7x-2)}} \right. \\\\
 \left \{ {{3x-4kx=4k} \atop {4x-7kx=-2k-4}} \right. \\\\
 \left \{ {{x(3-4k)=4k} \atop {x(4-7k)=-2k-4}} \right. \\\\
 \left \{ {{x=\dfrac{4k}{3-4k} \atop {x=\dfrac{-2k-4}{4-7k} \right. \\\\

\dfrac{4k}{3-4k}=\dfrac{-2k-4}{4-7k}\\\\

36k^2-6k-12=0\\
6k^2-k-2=0\\

\boxed{k= \dfrac{2}{3} \ or\ k= -\dfrac{1}{2}}\\\\

x=\dfrac{4*\dfrac{2}{3} }{3-4*\dfrac{2}{3} }=8\\
x=\dfrac{4*\dfrac{-1}{2} }{3-4*\dfrac{-1}{2} }=-\dfrac{2}{5}\\





$

$\boxed{x= 8 \ or\ x= -\dfrac{2}{5}}\\\\$

5 0

3 years ago