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2 years ago

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2.375 divided by 0.05

1 answer:

Xelga [282]2 years ago

4 0

Well, you calculate 237.5 divided by 5. This is 47.5.

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Is 5(x+1) the same as 5x+1? Explain your answer.

Morgarella [4.7K]

No
5*x= x (correct)
5*1=5 (incorrect)
When I say correct or incorrect it means that it does/doesn’t match 5x+1

6 0

2 years ago

The area of a rectangular pool is 6365 m2.<br> If the length of the pool is 95 m, what is its width?

PilotLPTM [1.2K]

Answer:

67

6365/95 = 67

But if your counting METERS and SQUARE METERS

It would be 134

3 0

2 years ago

four families go to the cinema. the reid family take 2 adults and 2 children for £18 ; the mghee family take 1 senior, 2 adults

RoseWind [281]

Answer:

$\pounds 26.50$

Step-by-step explanation:

Let

x = cost for one adult

y = cost for one child

z = cost for one senior.

<u>The Reid family:</u>

take 2 adults and 2 children and paid $2x+2y$ that is $\pounds 18$ , so

$2x+2y=18$

<u>The Mghee family:</u>

take 1 senior, 2 adults and 1 child and paid $x+2x+y$ that is $\pounds 18.50$ , so

$x+2x+y=18.50$

<u>The Griffiths family:</u>

take 1 senior and 3 adults and paid $z+3x$ that is $\pounds 19.50$ , so

$z+3x=19.50$

You get the system of three equations:

$\left\{\begin{array}{l}2x+2y=18\\ \\z+2x+y=18.50\\ \\z+3x=19.50\end{array}\right.$

From the first equation:

$2y=18-2x\\ \\y=9-x$

From the third equation:

$z=19.50-3x$

Substitute them into the second equation:

$19.50-3x+2x+9-x=18.50\\ \\-3x+2x-x=18.50-19.50-9\\ \\-2x=-10\\ \\2x=10\\ \\x=5$

Then

$y=9-5=4\\ \\z=19.50-3\cdot 5=19.50-15=4.50$

Hence,

the cost for one adult is $\pounds 5$

the cost for one child is $\pounds 4$

the cost for one senior is $\pounds 4.50$

<u>The Linton family</u> takes 1 senior, 2 adults and 3 children and paid

$\pounds 4.50+2\cdot \pounds 5+3\cdot \pounds 4=\pounds 26.50$

3 0

2 years ago

Which relation is a function?

Leno4ka [110]

Answer:

The bottom left one

Step-by-step explanation:

They are the only graph that has one output for each x value

3 0

1 year ago

G is the incenter, or point of concurrency, of the angle bisectors of ΔACE.

vova2212 [387]

Answer:

▢ $\overline{DG}$ ≅ $\overline{FG}$

▢ $\overline{DG}$ ≅ $\overline{BG}$

▢ $\overline{GE}$ bisects $\displaystyle ∠DEF$

▢ $\overline{GA}$ bisects $\displaystyle ∠BAF$

Step-by-step explanation:

As you can see, angles $\displaystyle BAF$ and $\displaystyle DEF$ form two half-kites, which you should know have <em>two</em><em> </em><em>pairs of</em><em> </em><em>congruent</em><em> </em><em>sides</em>, therefore segments $\displaystyle GA$ and $\displaystyle GE$ are what you call angle bisectours, which is why segments $\displaystyle FG,$ $\displaystyle DG,$ and $\displaystyle BG$ are congruent.

I am joyous to assist you at any time.

4 0

2 years ago