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3 years ago

7,676 rounded to the nearest hundred

1 answer:

7 0

the nearest hundred spot is 700 for the number. when rounding for hundreds, you use the 2nd digit number. Example, if a proablem asks to round 521 to the nearest hundred it would be 500. Another rule to remember is that anything 5 and above you round up (ex: 589 would round up to 600). Anything below 5 you would round down (ex: 234 would round down to 200). goinf back to the problem, the answer should be 7,700

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Factorize using middle term: m²+3m+2

Rina8888 [55]

Answer:

Step-by-step explanation:

m2 + 3m + 2

m2 + m + 2m + 2

m (m + 1) + 2(m + 1)

(m + 1) and (m + 2) are the factors

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3 years ago

An above swimming pool is leaking after 1/2 hour the pool has leaked 7/8 of a gallon of water

Luden [163]

The answer to your question are 1.75

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A jar contains quarters (worth $\$0.25$ each), nickels (worth $\$0.05$ each) and pennies (worth $\$0.01$ each). The value of the

denis-greek [22]

Answer:It doesn't make sense when I try to do it,I even tried a calculator

Step-by-step explanation: IDK?

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3 years ago

Read 2 more answers

Ab=45 mm,bc=10cm 4mm ,cd = 5 cm 6 mm ,de= 35 mm,ea=x cm, and the perimeter of irregular pentagon is 280 mm. Find the length of t

DaniilM [7]

Answer:

The length of the unknown side x is 4 cm

Step-by-step explanation:

we know that

The perimeter of a polygon is equal to the sum of its length sides

Remember that

1 cm=10 mm

In this problem we have

AB=45 mm ------> convert to cm -----> AB=45/10=4.5 cm

BC=10 cm 4 mm -----> convert to cm -----> BC=10+4/10=10.4 cm

CD=5 cm 6 mm -----> convert to cm -----> CD=5+6/10=5.6 cm

DE=35 mm ------> convert to cm -----> DE=35/10=3.5 cm

EA=x cm

P=280 mm ------> convert to cm -----> P=280/10=28 cm

The perimeter of the irregular polygon is equal to

P=AB+BC+CD+DE+EA

substitute the values

28=4.5+10.4+5.6+3.5+x

28=24+x

x=28-24

x=4 cm

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3 years ago

The functions f (theta) and g (theta) are sine functions, where f (0) equals g (0) equals 0.The amplitude of f (theta) is twice

777dan777 [17]

If period of $f(\theta)$ is one-half the period of $g(\theta)$ and
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To find the period of sine function $f(\theta)=asin(b\theta+c)$ we use the rule $T_{f}= \frac{2\pi}{b}$ .

f is sine function where f (0)=0, then c=0; with period $\pi$ , then $f(\theta)=asin 2\theta$ , because $T_{f}= \frac{2 \pi }{2} = \pi$ .

To find a we consider the condition $f( \frac{ \pi }{4} )=4$ , from where $asin2* \frac{\pi}{4} =a*sin \frac{ \pi }{2} =a=4$ .

If the amplitude of $f(\theta)$ is twice the amplitude of $g(\theta)$ , then $g(\theta)$ has a product factor twice smaller than $f(\theta)$ and while period of $g(\theta)$ is 2π and g(0)=0, we can write $g(\theta)=2sin\theta$ .

8 0

3 years ago