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2 years ago

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A customer purchased a total of 23 hamburgers and chicken sandwiches. Each hamburger costs $3.50 and each chicken sandwich costs

$4.50. If the customer spend a total of $88.50, how many hamburgers did the customer purchase?
Someone please answer I have a time limit

1 answer:

Bezzdna [24]2 years ago

6 0

Answer:

25 2/7

Step-by-step explanation:

88.50/3.50 divide 88.50 by 3:50

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What is the slope of the line represented by the equation y= 4/5x-3

guajiro [1.7K]

Answer:

m=

$\frac{4}{5}$

Step-by-step explanation:

Slope of a line is the co efficient of x in the given line.

$y = \frac{4}{5} x - 3 \\ y = mx + b$

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2 years ago

Marge purchased bicycle helmets and tire pumps. Each helmet cost $12.00 and each pump cost $8.00. She purchased a total of 18 it

algol [13]

Answer:

11

Step-by-step explanation:

Let the no. of helmet be x

cost of 1 helmet = $12.00

cost of x helmet = $12.00*x = $12x

Let the no. of tire pumps be y

cost of 1 tire pumps = $8.00

cost of x tire pumps = $8.00*y = $8y

Given that total no. of helmet and pump is 18

thus

x + y = 18

y = 18-x

also given

total money spent is $188

thus

12x+8y = 188

using y = 18 - x

we have

12x + 8(18-x) = 188

=> 12x+ 144 - 8x = 188

=> 4x = 188-144 = 44

=> x = 44/4 = 11

Thus, no of helmet bought by Margo is 11.

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2 years ago

Solve V = 1/3(3.14)r^2h for h

Ber [7]

<span>V = 1/3(3.14)r^2h
=1.046r^2h

Solving for V </span>

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2 years ago

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Es 18 de junio de 1815, las tropas napoleónicas se encuentran justo adelante, tu eres el ingeniero en balística y el general Wel

Airida [17]

Answer:

89.1° or -1.4°

Step-by-step explanation:

1. Location:

You are on the Mont-Saint-Jean escarpment, near the Belgian town of Waterloo.

The French troops are about 50 m below you and 1.2 km distant.

2. Finding the firing angle

Data:

R = 1200 m

u = 600 m/s

h = -50 m (the height of the target)

a = 9.8 m/s²

We have two conditions.

Horizontal distance

(1) 1200 = 600t cosθ

Vertical distance

(2) -50 = 600t sinθ - 4.9t²

Divide each side of (1) by 600cosθ.

$(3) \, t =\dfrac{2}{\cos \theta}$

Substitute (3) into (2)

$-50 = 600t \sin \theta - 4.9t^{2} = 600 \left( \dfrac{2}{\cos \theta} \right ) \sin \theta - 4.9 \left( \dfrac{2}{\cos \theta} \right )^{2}\\\\(4) \, -50 = 1200 \tan \theta - \dfrac{19.6}{\cos^{2} \theta}$

Recall that

(5) sec²θ = 1/cos²θ = tan²θ + 1

Substitute (5) into (4)

$-50 = 1200 \tan \theta - 19.6 \left(\tan^{2} \theta}+ 1\right )$

Set up a quadratic equation

$\begin{array}{rcl}-50 & = & 1200 \tan \theta - 19.6\tan^{2} \theta -19.6 \\0 & = & 1200 \tan \theta - 19.6\tan^{2} \theta + 30.4\\0 & =&19.6\tan^{2} \theta - 1200 \tan \theta - 30.4\\0 & =&\tan^{2} \theta - 61.224 \tan \theta - 1.551\\\end{array}$

Solve for θ

Use the quadratic formula.

tanθ = 61.249 or -0.025

θ = arctan(61.249) = 89.1° or

θ = arctan(-0.025) = -1.4°

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I don't Clearly understand your question but I think it is 0.428571

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