Pleas helppp its due soon

Mathematics

2 answers:

gulaghasi [49]3 years ago

5 0

Answer:

The last option, it is greater than the distance from the Movie Theater to the Arcade

Step-by-step explanation:

Distance from food court to store is 6 units

Distance from movie theater to arcade is 5 units

Usimov [2.4K]3 years ago

4 0

Answer:

Step-by-step explanation:

I dont think this needs explained

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The price of an item has been reduced by 70% the original price was $58 what’s the price of the item now ¿

Mamont248 [21]

Answer:$17.40

Step-by-step explanation:

This is a very simple question. First since 70% is the amount of money removed, we do 100-70=30

30% is the price that remains. Change that into a fraction 30/100 and then multiply the original cost

30/100*58=17.4

The answer is $17.40

6 0

3 years ago

Alan says that the distance from Hadley to Greenville is 16 miles. Is he correct?Explain

yuradex [85]

Answer:

Step-by-step explanation:

Considering the complete question attached in figure.

Consider

Hadley = x

Parker = y

Greenville = z

Distance between Greenville and Parker = yz = 29

Given that distance between Hadley and Greenville is 13 miles less than the distance between Parker and Greenville.

xz = yz -13 --- (1)

Distance between hardley and greenville = xz = ?

From (1)

xz = 29 -13

xz = 16 miles

So Alan is right that distance between Hadley and Greenville is 16 miles

8 0

3 years ago

A ball is thrown from an initial height of 1 meter with an initial upward velocity of 15m/s. The ball's height h (in meters) aft

lakkis [162]

$\bf \stackrel{\textit{ball's height}~\hfill }{\stackrel{\downarrow }{h}=1+15t-5t^2}\implies \stackrel{\textit{ball's height}~\hfill }{\stackrel{\downarrow }{6}=1+15t-5t^2}\implies 0=-5+15t-5t^2 \\\\\\ ~~~~~~~~~~~~\textit{using the quadratic formula} \\\\ \stackrel{\stackrel{a}{\downarrow }}{5}t^2\stackrel{\stackrel{b}{\downarrow }}{-15}t\stackrel{\stackrel{c}{\downarrow }}{+5}=0 \qquad \qquad t= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a}$

$\bf t=\cfrac{-(-15)\pm\sqrt{(-15)^2-4(5)(5)}}{2(5)}\implies t=\cfrac{15\pm\sqrt{225-100}}{10} \\\\\\ t=\cfrac{15\pm\sqrt{125}}{10}\implies t=\cfrac{15\pm\sqrt{5^2 \cdot 5}}{10}\implies t=\cfrac{\stackrel{3}{~~\begin{matrix} 15 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}\pm ~~\begin{matrix} 5 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~\sqrt{5}}{\underset{2}{~~\begin{matrix} 10 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}$