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3 years ago

Could someone please help and explain?

Mathematics

1 answer:

NeTakaya3 years ago

8 0

Answer:

Step-by-step explanation:

Basically, a number which is divisible by 4 formed by the last two digits in a number is divisible by 4. So, we can see 28 is divisible by 4... Therefore the anser is simply "2".

328÷4=82. :)) Hope it helps!

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Find BC if AC = 13 and AB = 10

maxonik [38]

Answer:

BC=3

Step-by-step explanation:

We can solve this by eliminating given detail of A.

AC=13 , AB=10

To find BC, We minus AC With AB

BC= AC-AB

BC=3

3 Is the final answer

I hope this helps, dont hesitate to ask for any question.

Mark me as brainliest is appreciated.Tq!!

7 0

3 years ago

Two cars are traveling down the highway with the same speed. If the first car increases its speed by 1km/hr, and the other car d

Olegator [25]

Answer:

The speed of the cars is $32 \frac{km}{h}$

Step-by-step explanation:

First we must first have the clear concept that $speed=\frac{distance}{time}$ or $s=\frac{d}{t}$

Our question is the speed of the cars then the variable to clear will be s.

Let's raise the equation for each car taking into account that we have the following data:

Car 1: $s_{1}=s+1$ , $t_{1}=2$ and $d_{1}= d_{2}$

Car 1: $s_{2}=s-10$ , $t_{2}=3$ and $d_{1}= d_{2}$

The two cars travel the same distance so we will raise the distance formula for each car and then match them.

Car 1

$d_{1}=s_{1}*t_{1}$

$d_{1}=(s+1)*2$

$d_{1}=2s+2$

Car 2

$d_{2}=s_{2}*t_{2}$

$d_{2}=(s-10)*3$

$d_{2}=3s.30$

$d_{1}=d_{2}$

$2s+2=3s-30$

$2+30=3s-2s$

$32=s$

The speed of the cars is 32 km/hr

7 0

3 years ago

write in scientific notation 1. 0.0000364 2. 0.00751 3. 0.10005 4. 1,094 5. 0.00000099 6. 0.04101 7. 10,500 8. 8,9000

solmaris [256]

$1.\\0\underbrace{.00003}_{5\to}54=3.54\times10^{-5}\\2.\\0\underbrace{.007}_{3\to}51=7.51\times10^{-3}\\3.\\0\underbrace{.1}_{1\to}0005=1.0005\times10^{-1}\\4.\\1\underbrace{,094}_{\leftarrow3}=1.094\times10^3\\5.\\0\underbrace{.0000009}_{7\to}9=9.9\times10^{-7}$
$6.\\0\underbrace{.04}_{2\to}101=4.101\times10^{-2}\\7.\\1\underbrace{0,500}_{\leftarrow4}=1.05\times10^4\\8.\\8\underbrace{9000}_{\leftarrow4}=8.9\times10^4$

3 0

3 years ago

Bus A and Bus B leave the bus depot at 7 am.

Art [367]

Answer:

9:30 am

Step-by-step explanation:

We can make a chart and keep filling it in until we see the same time in both columns.

Bus A Bus B

7:00 7:00

7:30 7:50

8:00 8:40

8:30 9:30

9:00 10:20

9:30

Answer: 9:30 am

We can also use the least common multiple of 30 and 50.

We need to find the smallest number that is a multiple of both 30 and 50.

30 = 2 * 3 * 5

50 = 2 * 5^2

LCD = 2 * 3 * 5^2 = 150

The first time the buses will be at the depot together after 7:00 am is 150 minutes later. 150 minutes = 2 hours 30 minutes.

7:00 am + 2 hours and 30 minutes = 9:30 am

Answer: 9:30 am

3 0

3 years ago

Read 2 more answers

Esi says that the value of z when z^2 = 80 does not belong in either of these categories since z must be greater than 8.

Anettt [7]

Answer:

Step-by-step explanation:

if z=8 then z^2 would be 64, which is not equal to 80

iz z=9 then z^2 would be 81, which is closer to 80 than 64

therefore, 9 is the whole number than z would be closest to

7 0

2 years ago

Read 2 more answers