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3 years ago

10

PLEASE PLEASE PLEASE HELP AND QUICK

1 answer:

prohojiy [21]3 years ago

7 0

Add all of the water he drank together:

$8 \frac{2}{8} + 8 \frac{4}{8} + 8 \frac{4}{8} + 8 \frac{4}{8} + 8 \frac{6}{8}$

Add fractions:

$\frac{2}{8} + \frac{4}{8} + \frac{4}{8} + \frac{4}{8} + \frac{6}{8} = \frac{20}{8} = 2 \frac{1}{2}$

Add whole numbers and fraction:

$8 + 8 + 8 + 8 + 8 + 2 \frac{1}{2} = 42 \frac{1}{2}$

Divide by 5:
42.5/5=8.5
Jorge should drink 8.5 water a day

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Determine which numbers the equations need to be multiplied by to form opposite terms of the y-variable. 3x - 1 4 y = 15 2 3 x -

sesenic [268]

In the elimination method, one variable is eliminated from the system of equations to find the value of the other.

To create the opposite terms of the x-variable-

The first equation be multiplied by the number 4.

Second equation be multiplied by the number -6.

<h3>What is the elimination method?</h3>

To solve the system of equations and find the value of the variables, the elimination method is used.

In this method, one variable is eliminated from the system of equations to find the value of the other.

Given information-

The system of the equation given in the problem is,

$\rm 3x-\dfrac{1}{4}y=15$

The second equation given in the problem is;

$\rm \dfrac{2}{3}x-\dfrac{1}{6}y=6$

To eliminate the x term, make the coefficient of x of both equations equal.

It can be done by multiplying the coefficient of x term of the first equation to the second equation and the coefficient of x term of the second equation to the first equation.

The first equation is multiplied by;

$\rm 4\times 3x-4\times \dfrac{1}{4}y=15\times 4\\\\12x-y=50$

The second equation is multiplied by;

$\rm -6 \times \dfrac{2}{3}x- -6 \times\dfrac{1}{6}y= -6 \times6\\\\-4x-y=-36\\\\4x+y=36$

Hence, to create the opposite terms of the x-variable-

The first equation is multiplied by the number 4.

The second equation is multiplied by the number -6.

Learn more about the elimination method here;

brainly.com/question/7013345

6 0

2 years ago

Read 2 more answers

Triangle JKL has vertices J(2,5), K(1,1), and L(5,2). Triangle QNP has vertices Q(-4,4), N(-3,0), and P(-7,1). Is (triangle)JKL

Tems11 [23]

Answer:

Yes they are

Step-by-step explanation:

In the triangle JKL, the sides can be calculated as following:

J(2;5); K(1;1)

=> JK = $\sqrt{(1-2)^{2} + (1-5)^{2} } = \sqrt{(-1)^{2}+(-4)^{2} } = \sqrt{1+16}=\sqrt{17}$

J(2;5); L(5;2)

=> JL = $\sqrt{(5-2)^{2} + (2-5)^{2} } = \sqrt{3^{2}+(-3)^{2} } = \sqrt{9+9}=\sqrt{18} = 3\sqrt{2}$

K(1;1); L(5;2)

=> KL = $\sqrt{(5-1)^{2} + (2-1)^{2} } = \sqrt{4^{2}+1^{2} } = \sqrt{1+16}=\sqrt{17}$

In the triangle QNP, the sides can be calculate as following:

Q(-4;4); N(-3;0)

=> QN = $\sqrt{[-3-(-4)]^{2} + (0-4)^{2} } = \sqrt{1^{2}+(-4)^{2} } = \sqrt{1+16}=\sqrt{17}$

Q (-4;4); P(-7;1)

=> QP = $\sqrt{[-7-(-4)]^{2} + (1-4)^{2} } = \sqrt{(-3)^{2}+(-3)^{2} } = \sqrt{9+9}=\sqrt{18} = 3\sqrt{2}$

N(-3;0); P(-7;1)

=> NP = $\sqrt{[-7-(-3)]^{2} + (1-0)^{2} } = \sqrt{(-4)^{2}+1^{2} } = \sqrt{16+1}=\sqrt{17}$

It can be seen that QPN and JKL have: JK = QN; JL = QP; KL = NP

=> They are congruent triangles

7 0

3 years ago

Read 2 more answers

Need help plsss ASAP

Phantasy [73]

Answer: I would choose option "c"

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Work out the height of a triangle with base b= 7.7mm and area a=164.78mm2

solong [7]

Answer:

21.4 mm

Step-by-step explanation:

area ÷ base = height

164.78 mm^2 ÷ 7.7 mm

21.4 mm

Check:

base • height = area

7.7 mm • 21.4 mm = 164.78 mm^2

5 0

3 years ago

Vichy reads at a constant rate. She can read 5 pages

vagabundo [1.1K]

Answer:

p = (5t)/9

Step-by-step explanation:

9 mins ---------- 5 pages

1 min -------------5/9 page

t mins ------------5/9 x t pages

Thus

p = (5t)/9

7 0

3 years ago