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3 years ago

22) John ordered a large pizza for $19.50. How much

Mathematics

2 answers:

Pani-rosa [81]3 years ago

8 0

Answer:2.34

Step-by-step explanation:

siniylev [52]3 years ago

7 0

i believe its $21.84 because 12% would be $2.34

Step-by-step explanation:

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Which number line model represents the expression -2/5 + 4/5

Akimi4 [234]

-2/5 + 4/5 = 2/5..................

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3 years ago

Write an equation in slope intercept form with the given information. Goes through points (0,3) and (1,6)

Setler79 [48]

Answer:

below..

Step-by-step explanation:

slope = (y - y1)/(x-x1) = (6 - 3 )/(1-0) = 3 is the slope/ gradient

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2 years ago

Guys!!! I need help ASAP!!!!

loris [4]

<h2><u>Solution 1</u> :</h2>

Cost of $1\frac{1}{3}$ pounds of gummy bears = $4.40

Cost of 1 pound of gummy bears :

$= 4.40 \div 1 \frac{1}{3}$

$= \frac{440}{100} \div \frac{4}{3}$

$= \frac{440}{100} \times \frac{3}{4}$

$= \frac{1320}{400}$

$= \frac{1320 \div 20}{400 \div 20}$

$= \frac{66}{20}$

$= \frac{66 \div 2}{20 \div 2}$

$= \frac{33}{ 10 }$

$= 3 \frac{3}{10}$

$= \color{plum}\bold{\$3.3}$

Therefore, 1 pound of gummy bears = <u>$3.3</u>

<h2><u>Solution 2</u> : </h2>

Cost of 1 pound of gummy bears = $3.3

Money Daniel has = $1.00

Quantity of gummy bears he can afford :

$= 1 \div 3.3$

$= 1 \div 3 \frac{3}{10}$

$= 1 \div \frac{33}{10}$

$= 1 \times \frac{10}{33}$

$= \frac{10}{33}$

$= 0.30$

Therefore, Daniel will be able to buy <u>0.30 pounds</u> of gummy bears from Jio.

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2 years ago

An airplane descends 2.2 miles to an elevation of 6.75 miles find the elevation of the plane before its descent

mrs_skeptik [129]

Answer: 8.75 miles

Explanation: this is simple, just add the descending variable to the current elevation to get the previous elevation.

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3 years ago

Read 2 more answers

How to work out this linear stimultaneous equations

Oduvanchick [21]

Step-by-step explanation:

extract the third equation from one of the given equation by making at most one variable with 1 as the coefficient the subject of the equation.
$x = 2 - y...........(3)$
now substitute the third equation to the other equation *Not the one you extracted the new equation from*
$y = 2 ({2 - y})^{2} + (2 -y ) - 2$
After substituting simplify
$y = 2(2 - y)(2 - y) + 2 - y - 2 \\ y = 2(4 - 2y - 2y + {y}^{2} ) + 2 - y - 2 \\ y = 2(4 - 4y + {y}^{2} ) - y \\ y = 8 - 8y + 2 {y }^{2} - y$
Therfore group the terms in the form of a general Quadratic equation.

7 0

1 year ago