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3 years ago

A square has side lengths of x. A rectangle is five

Mathematics

1 answer:

evablogger [386]3 years ago

7 0

Answer:

Ty for 50 point i love you muah muah

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How do I solve 2x+3y=7, x-y=1

Fittoniya [83]

<span>2x+3y=7
x-y=1

</span>2x+3y=7
x=1+y

2*(1+y)+3y=7
x=1+y

2+2y+3y=7
x=1+y

5y=5 |(:5)
x=1+y

y=1
x=2

CHECK:
2x+3y=7
x-y=1

2*2 + 3*1 = 7

x-y = 1
2-1 = 1

5 0

3 years ago

Read 2 more answers

A pitcher has 16 cups of water in it. During the day, Aditi drank 5⁄2 cups, Kavitha drank 15⁄4 cups, and Rahul drank 33⁄8 cups.

iragen [17]

Answer:

5/2 = 2 1/2

15/4 = 3 3/4

33 /8 = 4 1/8

Step-by-step explanation:

5/2 2 goes into 5 2 times with 1 left over = 2 1/2

15 /4 4 goes into 15 3 times with 3 left over 3 3/4

33/8 8 goes into 33 4 times with 1 left over 4 1/8

5 0

3 years ago

Write an equation in the slope-intercept form of the line.

nata0808 [166]

Answer: Y= 1/-2x + 2

4 0

2 years ago

Can someone please help me solve this?

cestrela7 [59]

Answer:

(a)

Step-by-step explanation:

The tangent and the normal at point P are perpendicular.

Given

$m_{tangent}$ = 3, then

Given the gradient m of the tangent then the gradient of a line perpendicular to it is

$m_{normal}$ = - $\frac{1}{m}$ = - $\frac{1}{3}$ → (a)

3 0

3 years ago

Read 2 more answers

I need help.. i really want to go sleep.. thank you so much...

Effectus [21]

Answer:

1) True 2) False

Step-by-step explanation:

1) Given $\sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i}$

To verify that the above equality is true or false:

Now find $\sum\limits_{k=0}^8\frac{1}{k+3}$

Expanding the summation we get

$\sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{0+3}+\frac{1}{1+3}+\frac{1}{2+3}+\frac{1}{3+3}+\frac{1}{4+3}+\frac{1}{5+3}+\frac{1}{6+3}+\frac{1}{7+3}+\frac{1}{8+3}$ $\sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}$

Now find $\sum\limits_{i=3}^{11}\frac{1}{i}$

Expanding the summation we get

$\sum\limits_{i=3}^{11}\frac{1}{i}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}$

Comparing the two series we get,

$\sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i}$ so the given equality is true.

2) Given $\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\sum\limits_{i=1}^3\frac{3i}{i+5}$

Verify the above equality is true or false

Now find $\sum\limits_{k=0}^4\frac{3k+3}{k+6}$

Expanding the summation we get

$\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\frac{3(0)+3}{0+6}+\frac{3(1)+3}{1+6}+\frac{3(2)+3}{2+6}+\frac{3(3)+4}{3+6}+\frac{3(4)+3}{4+6}$

$\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\frac{3}{6}+\frac{6}{7}+\frac{9}{8}+\frac{12}{8}+\frac{15}{10}$

now find $\sum\limits_{i=1}^3\frac{3i}{i+5}$

Expanding the summation we get

$\sum\limits_{i=1}^3\frac{3i}{i+5}=\frac{3(0)}{0+5}+\frac{3(1)}{1+5}+\frac{3(2)}{2+5}+\frac{3(3)}{3+5}$

$\sum\limits_{i=1}^3\frac{3i}{i+5}=\frac{3}{6}+\frac{6}{7}+\frac{9}{8}$

Comparing the series we get that the given equality is false.

ie, $\sum\limits_{k=0}^4\frac{3k+3}{k+6}\neq\sum\limits_{i=1}^3\frac{3i}{i+5}$

6 0

3 years ago