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Ede4ka [16]
3 years ago
10

rachel buys 1,5 pounds of grapes. She eats 0,3 of that amount on Tuesday and 0,2 of amount on Wednesday. How many pounds of grap

es are left
Mathematics
1 answer:
Alisiya [41]3 years ago
7 0

Answer:

Can you show how you got this

Step-by-step explanation:

You might be interested in
Please help me if you can ‼️
Karolina [17]

Answer:

y=14.5x+5

Step-by-step explanation:

Let

x ----> the number of tickets

y ----> the price of tickets

we have the ordered pairs

(4,63) and (6,92)

step 1

<em>Find the slope of the linear equation</em>

The formula to calculate the slope between two points is equal to

m=\frac{y2-y1}{x2-x1}

substitute the values

m=\frac{92-63}{6-4}

m=\frac{29}{2}=\$14.5\ per\ ticket

step 2

Find the y-intercept or initial value of the linear equation

we know that

The linear equation in slope intercept form is equal to

y=mx+b

where

m is the slope b is the y-intercept

we have

m=14.5

point\ (4,63)

substitute

63=14.5(4)+b

solve for b

b=63-14.5(4)

b=\$5

In this context the y-intercept is a one charge fee for the ticket service

The equation is equal to

y=14.5x+5

4 0
3 years ago
P(x) = x + 1x² – 34x + 343<br> d(x)= x + 9
Feliz [49]

Answer:

x=\frac{9}{d-1},\:P=\frac{-297d+378}{\left(d-1\right)^2}+343

Step-by-step explanation:

Let us start by isolating x for dx = x + 9.

dx - x = x + 9 - x > dx - x = 9.

Factor out the common term of x > x(d - 1) = 9.

Now divide both sides by d - 1 > \frac{x\left(d-1\right)}{d-1}=\frac{9}{d-1};\quad \:d\ne \:1. Go ahead and simplify.

x=\frac{9}{d-1};\quad \:d\ne \:1.

Now, \mathrm{For\:}P=x+1x^2-34x+343, \mathrm{Subsititute\:}x=\frac{9}{d-1}.

P=\frac{9}{d-1}+1\cdot \left(\frac{9}{d-1}\right)^2-34\cdot \frac{9}{d-1}+343.

Group the like terms... 1\cdot \left(\frac{9}{d-1}\right)^2+\frac{9}{d-1}-34\cdot \frac{9}{d-1}+343.

\mathrm{Add\:similar\:elements:}\:\frac{9}{d-1}-34\cdot \frac{9}{d-1}=-33\cdot \frac{9}{d-1} > 1\cdot \left(\frac{9}{d-1}\right)^2-33\cdot \frac{9}{d-1}+343.

Now for 1\cdot \left(\frac{9}{d-1}\right)^2 > \mathrm{Apply\:exponent\:rule}: \left(\frac{a}{b}\right)^c=\frac{a^c}{b^c} > \frac{9^2}{\left(d-1\right)^2} = 1\cdot \frac{9^2}{\left(d-1\right)^2}.

\mathrm{Multiply:}\:1\cdot \frac{9^2}{\left(d-1\right)^2}=\frac{9^2}{\left(d-1\right)^2}.

Now for 33\cdot \frac{9}{d-1} > \mathrm{Multiply\:fractions}: \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c} > \frac{9\cdot \:33}{d-1} > \frac{297}{d-1}.

Thus we then get \frac{9^2}{\left(d-1\right)^2}-\frac{297}{d-1}+343.

Now we want to combine fractions. \frac{9^2}{\left(d-1\right)^2}-\frac{297}{d-1}.

\mathrm{Compute\:an\:expression\:comprised\:of\:factors\:that\:appear\:either\:in\:}\left(d-1\right)^2\mathrm{\:or\:}d-1 > This\: is \:the\:LCM > \left(d-1\right)^2

\mathrm{For}\:\frac{297}{d-1}:\:\mathrm{multiply\:the\:denominator\:and\:numerator\:by\:}\:d-1 > \frac{297}{d-1}=\frac{297\left(d-1\right)}{\left(d-1\right)\left(d-1\right)}=\frac{297\left(d-1\right)}{\left(d-1\right)^2}

\frac{9^2}{\left(d-1\right)^2}-\frac{297\left(d-1\right)}{\left(d-1\right)^2} > \mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}> \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}

\frac{9^2-297\left(d-1\right)}{\left(d-1\right)^2} > 9^2=81 > \frac{81-297\left(d-1\right)}{\left(d-1\right)^2}.

Expand 81-297\left(d-1\right) > -297\left(d-1\right) > \mathrm{Apply\:the\:distributive\:law}: \:a\left(b-c\right)=ab-ac.

-297d-\left(-297\right)\cdot \:1 > \mathrm{Apply\:minus-plus\:rules} > -\left(-a\right)=a > -297d+297\cdot \:1.

\mathrm{Multiply\:the\:numbers:}\:297\cdot \:1=297 > -297d+297 > 81-297d+297 > \mathrm{Add\:the\:numbers:}\:81+297=378 > -297d+378 > \frac{-297d+378}{\left(d-1\right)^2}

Therefore P=\frac{-297d+378}{\left(d-1\right)^2}+343.

Hope this helps!

5 0
4 years ago
You have two cards with a sum of (-12) in your hand. what two cards could you hsve
kirill115 [55]

The way the question is written, you could have almost anything in your hand. All they have to do is add to - 12.

-14 and 2

- 11 and - 1

-12 and 0

- 82 and 70

4 0
4 years ago
Can you solve this and don't respond that you don't know <br><br><br><br><br><br><br><br><br><br>​
Fofino [41]

The answer is going to be

B)

1/8 Lb.

6 0
3 years ago
Read 2 more answers
I DONT KNOW THESE! HELP!
kkurt [141]

Answer:

20. x = 5, y = -2

21. x = 11, y = 12

22. x = 22, y = 11

23. x = 11, y = 10

Step-by-step explanation:

20. Opposite sides in the parallelogram are congruent, so

2y+18=3x-1\\ \\6x-3=17-5y

Solve this system of two equations:

\left\{\begin{array}{l}2y-3x=-19\\ \\6x+5y=20\end{array}\right.

Multiply the first equation by 2 and add two equations:

2(2y-3x)+6x+5y=2\cdot (-19)+20\\ \\4y-6x+6x+5y=-38+20\\ \\9y=-18\\ \\y=-2

Substitute it into the first equation:

2\cdot (-2)-3x=-19\\ \\-3x=-19+4\\ \\-3x=-15\\ \\x=5

21. Opposite angles in the parallelogram are congruent, so

11x+5=10y+6

Consecutive angles are supplementary, so

6x-y+11x+5=180^{\circ}

Solve this system of two equations:

\left\{\begin{array}{l}11x-10y=1\\ \\17x-y=175\end{array}\right.

From the second equation

y=17x-175

Substitute it into the first equation:

11x-10(17x-175)=1\\ \\11x-170x+1750=1\\ \\-159x=-1749\\ \\x=11\\ \\y=17\cdot 11-175=187-175=12

22. Opposite angles in the parallelogram are congruent, so

2x-5=3y-12

Consecutive angles are supplementary, so

2x-5+7y+x=180^{\circ}

Solve this system of two equations:

\left\{\begin{array}{l}2x-3y=-7\\ \\3x+7y=185\end{array}\right.

From the first equation

x=-3.5+1.5y

Substitute it into the second equation:

3(-3.5+1.5y)+7y=185\\ \\-10.5+4.5y+7y=185\\ \\-105+45y+70y=1,850\\ \\115y=1,850+105\\ \\115y=1,955\\ \\y=17\\ \\x=-3.5+1.5\cdot 17=22

23. Opposite sides in the parallelogram are congruent, so

2x+9=4y-9\\ \\3x-5=2y+8

Solve this system of two equations:

\left\{\begin{array}{l}2x-4y=-18\\ \\3x-2y=13\end{array}\right.

Multiply the second equation by 2 and subtract it from the first equation:

2x-4y-2(3x-2y)=-18-2\cdot 13\\ \\2x-4y-6x+4y=-18-26\\ \\-4x=-44\\ \\x=11

Substitute it into the first equation:

2\cdot 11-4y=-18\\ \\-4y=-18-22\\ \\-4y=-40\\ \\y=10

3 0
3 years ago
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