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4 years ago

58 Points & Brainliest I PROMISE!!!

2 answers:

7 0

$the \: \sqrt{49} = 7 \: and \: the \: \sqrt{64} = 8 \: the \: halfway \\ point \: of \sqrt{49} \: to \: \sqrt{64} \: would \: be \: \sqrt{56} . \: half \: of \\that \: woud \: be \: \sqrt{52} . \: so \: \sqrt{52} \: is \: a \: quaterway \\ to \: \sqrt{49} \: to \: the\: \sqrt{64}. \: estimating,\: it \: would \: be \\7.25 \: or \: 7 \frac{1}{4}$

Sergeeva-Olga [200]4 years ago

5 0

49 < 52 < 64

sqrt (49) < sqrt (52) < sqrt (64)

7 < sqrt ( 52) < 8

the sqrt(52) is close to 7

You might be interested in

Can some one help me on this plz

skad [1K]

Step 1
switch x and y
[x=-3y+27]

step 2
solve for y in steps
[x-27=-3y]

step 3 another solving
[(x-27)/-3=y]

step 4 simplify
[-1/3x+9=y]
(since -27/-3=+9)

attachment below

4 0

4 years ago

The area of Australia is approximately 7.7 x 10^6 km. If it were to be divided into states of about 1.54 x 10^6 km, how many sta

Goshia [24]

Answer:

7.7 x 10^6 / 1.54 x 10^6

note that the 10^6 / 10^6 cancles out (gets to *1 or rather /1)

7.7 / 1.54 is exactly 5

8 0

3 years ago

Graph the line with the equation y = 1/6x+3

Neporo4naja [7]

Step-by-step explanation:

There isnt even a graph shown to graph on lol

4 0

3 years ago

Set up a system of equations for the following scenario. Then solve for the system. Three students buy different combinations of

Natasha2012 [34]

Let
x = cost of a ticket for a senior
y = cost of a ticket for an adult
z = cost of a ticket for a student.

The first student buys 2 senior, 1 adult, and 2 student tickets for $51.
Therefore
2x + y + 2z = 51 (1)

Th second student buys 1 adult and 5 student tickets for $5.
Therefore
y + 5z = 55 (2)

The third student buys 2 senior, 2 adult, and 7 student tickets for $75.
Therefore
2x + 2y + 7z = 75 (3)

Answer:
The system of equation for determining x, y, and z is
2x + y + 2z = 51
y + 5z = 55
2x + 2y + 7z = 75

Warnng: The system of equations does not have a solution.

3 0

3 years ago

Plz Help???

GREYUIT [131]

1.
12 liters of 26% acid solution contain $26\% \times 12=0.26 \times 12=3.12$ liters of acid.
x liters of 40% acid solution contain $40\%x=0.4x$ liters of acid.
The third solution is made from the two mixed solution so its volume is $12+x$ , and the volume of acid in it is $0.4x+3.12$ . Its concentration is 33% or 0.33.
The concentration of a solution is the volume of the solution divide by the volume of acid in it.

$\frac{0.4x+3.12}{12+x}=0.33 \\ 0.4x+3.12=0.33(12+x) \\ 0.4x+3.12=3.96+0.33x \\
0.07x=0.84 \\
x=12$

The answer is B.

2.
12 liters of 16% acid solution contain $16\% \times 12=0.16 \times 12=1.92$ liters of acid.
x liters of 40% acid solution contain $40\%x=0.4x$ liters of acid.
The volume of the third solution is $12+x$ , and the volume of acid in it is $0.4x+1.92$ . Its concentration is 32% or 0.32.

$\frac{0.4x+1.92}{12+x}=0.32 \\
0.4x+1.92=0.32(12+x) \\
0.4x+1.92=3.84+0.32x \\
0.08x=1.92 \\
x=24$

The answer is B.

3.
16 liters of 18% acidsolution contain $18\% \times 16=0.18 \times 16=2.88$ liters of acid.
x liters of 27% acid solution contain $27\%x=0.27x$ liters of acid.
The volume of the third solution is $16+x$ , and the volume of acid in it is $0.27x+2.88$ . Its concentration is 21% or 0.21.

$\frac{0.27x+2.88}{16+x}=0.21 \\
0.27x+2.88=0.21(16+x) \\
0.27x+2.88=3.36+0.21x \\
0.06x=0.48 \\
x=8$

The answer is A.

5 0

4 years ago