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3 years ago

How many times can 10 go into 44

Mathematics

2 answers:

Nitella [24]3 years ago

5 0

4 times because if you count 10, 20, 30, 40, so it has to be 4

Masteriza [31]3 years ago

5 0

44/10 = 4.4
So 4.4 should be the answer.

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The groundkeep filled the first sandbox with sand to a height of 6 inches what is the volume in cubic feet of the sand that was

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Answer:

https://nrckids.org/CFOC/Infant_Toddlers

Step-by-step explanation:

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3 years ago

Given: rectangle ABCD is similar to rectangle ZBXY. If BC = 10, BX = 6, XY = 4, then CD =

maksim [4K]

Answer:

$CD=6\frac{2}{3}\ units$

Step-by-step explanation:

we know that

If rectangle ABCD is similar to rectangle ZBXY

then

the ratio of their corresponding sides is equal and is called the scale factor

$\frac{BC}{BX}=\frac{CD}{XY}$

in this problem we have

$BC=10\ units$

$BX=6\ units$

$XY=4\ units$

Substitute the values and solve for CD

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3 years ago

Read 2 more answers

The hypotenuse of a 45°-45°-90° triangle measures 4 cm. What is the length of one leg of the triangle? 2 cm 2 StartRoot 2 EndRoo

Oxana [17]

Answer:

$2\sqrt{2} cm$

Step-by-step explanation:

In order to solve this you just have to remember the formula for the sides and hypotenuse of the triangle rectangle:

$c^{2} =a^{2} +b^{2}$

Since A and B are equal we will use X to represent them both:

$c^{2} =a^{2} +b^{2}\\4^{2} =x^{2} +x^{2}\\16 =2x^{2}\\8=x^{2} \\x=\sqrt{8}=\sqrt{4*2}=2\sqrt{2}$

ANd by solving the equation we get that the length of the legs of the triangle rectangle with a hypotenuse of 4cm is $2\sqrt{2} cm$

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3 years ago

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Answer:

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3 years ago

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Wanni cycled 6 km from her house to the school at a uniform speed, v km/h. If she increased her speed

Karolina [17]

Answer:

The quadratic equation in terms of v is v² + 2 v + 180 = 0

Step-by-step explanation:

Given as :

The distance between house to the school = d = 6 km

The uniform speed = v km/h

So, Time = $\dfrac{\textrm Distance}{\textrm speed}$

or, t = $\dfrac{\textrm d}{\textrm v}$

Or, t = $\dfrac{\textrm 6}{\textrm v}$

<u>Now, Again</u>

The speed is increase by 2 km/h

i.e speed = (v + 2) km/h

So, Time taken = t' = (t - $\dfrac{4}{60}$ )hours

i.e t' = (t - $\dfrac{1}{15}$ )hours

Now, Time = $\dfrac{\textrm Distance}{\textrm speed}$

So, (t - $\dfrac{1}{15}$ ) = $\dfrac{\textrm d}{\textrm v}$

Or, (t - $\dfrac{1}{15}$ ) = $\dfrac{\textrm 6}{\textrm (v + 2)}$

Or , $\dfrac{\textrm 6}{\textrm v}$ - $\dfrac{1}{15}$ = $\dfrac{\textrm 6}{\textrm (v + 2)}$

Or , $\dfrac{\textrm 90 - v}{\textrm 15 v}$ = $\dfrac{\textrm 6}{\textrm v + 2}$

Or, (90 - v) × (v + 2) = 6 × 15 v

Or, 90 v - 180 - v² - 2 v = 90 v

Or, v² + 2 v + 180 = 90 v - 90 v

Or, v² + 2 v + 180 = 0

So, The quadratic equation in terms of v

v² + 2 v + 180 = 0

Hence The quadratic equation in terms of v is v² + 2 v + 180 = 0 Answer

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3 years ago