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Vilka [71]
3 years ago
7

I’ll give brainliest

Mathematics
1 answer:
nasty-shy [4]3 years ago
5 0

Answer:

A. 5 miles

B. 20 minutes

C. 12.5 miles per hour

You might be interested in
What is the value of this expression?<br><br> (−4)4
Marina86 [1]

Answer:

-16

Step-by-step explanation:

Anything written in the form of a(b) or (b)a means a × b:

(-4)4 means -4 × 4 which is -16 (a positive and a negative multiplied together make a negative).

Hope this helps!

8 0
3 years ago
For a certain map, the scale shows that 1 cm on the map equals 2.50 km on earth's surface. two towns on the map are 4.75 cm apar
Julli [10]
<span>In this problem, to find the answer we have to setup a series of ratios that relate the scale to real life distance. We know that 1cm = 2.50km, so that ratio would be 1cm/2.5km. For two towns that are 4.75cm apart on the map, we set a ration of 4.75cm/x km, where x is the actual distance. Now we set the ratios equal to each other and solve for x. 1/2.5=4.75/x where x = 4.75*2.5/1 = 11.875 and rounding up we get 11.88 km. The two towns are actually 11.88 km apart from each other.</span>
3 0
3 years ago
In 1997, the American Diabetes Association (ADA) and the federal government lowered the standard for diagnosing diabetes from a
Klio2033 [76]

Answer:

10 percent

Step-by-step explanation:

- Before 1997, people with a minimum FBGL of 140mg/dl were diagnosed as diabetic

- After 1997, people with a minimum FBGL of 126mg/dl were diagnosed as diabetic

- What is the proportion of people who were not considered diabetic before 1997 but are now considered diabetic (after 1997)?

140 - 126 = 14

In percentage, this proportion is = 14/140 × 100

= 10%

Hence, 10% of the people who were not considered diabetic before 1997 are now or will now be considered diabetic.

3 0
3 years ago
The diagram shows a triangle ABC where AB=AC , BC=AC, and ⦟BAC = 20 degree, find ⦟ADB​
Mariulka [41]

Answer:

hello something is wrong with the question. if ab= AC, and AC = BC, AB=BC, it'd make this an equilateral triangle and the degree wouldn't be 20 degrees

6 0
3 years ago
Hi guys, can anyone help me with this triple integral? Many thanks:)
Crank

Another triple integral.  We're integrating over the interior of the sphere

x^2+y^2+z^2=2^2

Let's do the outer integral over z.   z stays within the sphere so it goes from -2 to 2.

For the middle integral we have

y^2=4-x^2-z^2

x is the inner integral so at this point we conservatively say its zero.  That means y goes from -\sqrt{4-z^2} and +\sqrt{4-z^2}

Similarly the inner integral x goes between \pm-\sqrt{4-y^2-z^2}

So we rewrite the integral

\displaystyle \int_{-2}^{2} \int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}} \int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dx \; dy \; dz

Let's work on the inner one,

\displaystyle\int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dz

There's no z in the integrand, so we treat it as a constant.

=(x^2+xy+y^2)z \bigg|_{z=-\sqrt{4-y^2-z^2}}^{z=\sqrt{4-y^2-z^2}}

So the middle integral is

\displaystyle\int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}2(x^2+xy+y^2)\sqrt{4-y^2-z^2} \ dy  

I gotta go so I'll stop here, sorry.

7 0
3 years ago
Read 2 more answers
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