Which statement correctly describes the expression (-70)(-6)?

Mathematics

2 answers:

ddd [48]2 years ago

4 0

The answer is A because the factors are both negative

EastWind [94]2 years ago

3 0

Answer: the answer is letter A

Step-by-step explanation:

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L'shanda rode her bike 2.5 miles. Shaman rode her bike 3 miles. How many more feet did shanay rode her bike than l'shanda

9966 [12]

3 - 2.5 = 0.5

shanay rode 1/2 a mile more

hope this helps :)

6 0

2 years ago

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Beau buys a 3-pound bag of trail mix for a hike. He wants to make one-ounce bags for his friends with whom he is hiking. How man

den301095 [7]

1 pound = 16 ounces

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2 years ago

Helppppppp please, all math lovers I'm calling for you

Katen [24]

Answer:

Dive by n square root two, also option one.

Step-by-step explanation:

The first step to any algebra equation is to isolate the variable and you do that by divide square root 2 since it cancels out on the left side and now n is alone.

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2 years ago

The equation V = 16300 (0.94)^t represents the value (in dollars) of a car t years after its purchase. Use this equation to comp

Dafna1 [17]

Solution:

Given:

$V=16300(0.94)^t$

The value of a car after t - years will depreciate.

Hence, the equation given represents the value after depreciation over t-years.

To get the rate, we compare the equation with the depreciation formula.

$\begin{gathered} A=P(1-r)^t \\ \text{where;} \\ P\text{ is the original value} \\ r\text{ is the rate} \\ t\text{ is the time } \end{gathered}$

Hence,

$\begin{gathered} V=16300(0.94)^t \\ A=P(1-r)^t \\ \\ \text{Comparing both equations,} \\ P=16300 \\ 1-r=0.94 \\ 1-0.94=r \\ r=0.06 \\ To\text{ percentage,} \\ r=0.06\times100=6\text{ \%} \\ \\ \text{Hence, } \\ P\text{ is the purchase price} \\ r\text{ is the rate} \end{gathered}$

Therefore, the value of this car is decreasing at a rate of 6%. The purchase price of the car was $16,300.

5 0

2 months ago

Lnx=-3

scZoUnD [109]

Answer:

because It's rounded to the nearest hundred place

Step-by-step explanation:

Given equation: