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4 years ago

Help, I need the bottom three answered and I'm sorta stuck on four too. Not sure if it's right

Mathematics

1 answer:

amm18124 years ago

7 0

5. 15 yards = 13.716 meters, so 13.716 meters
6. 32.366 meters per second
7. 32÷2=16 and 16×60= 960, so it can eat 16 in one minute, and 960 in one hour.
Hope this helps!

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The midpoint of AB is M(1, -4). If the coordinates of A are (-3,-6), what are the coordinates of B?

kramer

Answer:

The coordinates of B is (5,-2)

Step-by-step explanation:

In order to find the coordinates of B, you have to use midpoint formula. Then substitute the following values into the formula :

$m = ( \frac{x1 + x2}{2} , \frac{y1 + y2}{2} )$

Let (x1,y1) be A (-3,-6),

Let (x2,y2) be B,

Let midpoint be M (1,-4),

$(1, - 4) = (\frac{ - 3 + x2}{2} , \frac{ - 6 + y2}{2} )$

$by \: comparison, \:$

$\frac{ - 3 + x2}{2} = 1$

$- 3 + x2 = 2$

$x2 = 5$

$\frac{ - 6 + y2}{2} = - 4$

$- 6 + y2 = - 8$

$y2 = - 2$

6 0

3 years ago

If the subject without diabetes eats a high-carb food such as the candy bar or orange smoothie, then the blood glucose levels wi

zysi [14]

Answer:

1. The right answer is stay the same.

In a healthy subject, a diet rich in sugar (such as starch and fast sugars) increases blood sugar in the first two hours after ingestion and then stabilizes after the third hour with the action of the patient. insulin which has a hypoglycemic action. But over time the pancreas (the organ that secretes insulin) will get tired over time if these foods are eaten excessively.

2. The right answer is stay the same.

In a healthy person, a diet of low-sugar foods (such as fiber) does not increase blood glucose (or very little) and helps maintain normal blood glucose levels. These foods are good for the health because they preserve the pancreas as long as there is not much solicitude by the presence of a hause of glucose in the blood.

3. The right answer is increase.

In a type 1 diabetic patient, the pancreas does not secrete insulin appropriately, and this is due to self-destruction of the organ by the immune system (autoimmune disease). This implies that during the ingestion of foods rich in sugars, the blood sugar will be hard to regain its normal level, that is why it remains elevated after three hours after ingestion.

4. The right answer is stay the same.

Foods low in carbohydrates are very good for the health and is recommended in the subjects diabetic (type 1 and 2) and even for the healthy subjects to avoid the appearance of the diabetes afterwards. These foods help diabetics to have controlled blood glucose levels when their degradation does not give glucose.

6 0

3 years ago

A^3+a+a^2+1 2a^2+2ab+ab^2+b^3

Vikentia [17]

$\frac{a^3+a+a^2+1}{a^3+a^2+ab^2+b^2} \cdot \frac{2a^2+2ab+ab^2+b^3}{a^3+a+a^2b+b} =\frac{a(a^2+1)+1(a^2+1)}{a^2(a+1)+b^2(a+1)} \cdot \frac{2a(a+b)+b^2(a+b)}{a(a^2+1)+b(a^2+1)} =\\\\=\frac{(a^2+1)(a+1)}{(a+1)(a^2+b^2)} \cdot \frac{(a+b)(2a+b^2)}{(a^2+1)(a+b)} = \frac{2a+b^2}{a^2+b^2}$

6 0

3 years ago

A train leaves little? rock, arkansas, and travels north at 80 kilometers per hour. another train leaves at the same time and tr

balandron [24]

p(80+60)=560

p(140)=560

560/140 = 4 hours

4 0

3 years ago

A couple intends to have two children, and suppose that approximately 52% of births are male and 48% are female.

Pachacha [2.7K]

a) Probability of both being males is 27%

b) Probability of both being females is 23%

c) Probability of having exactly one male and one female is 50%

Step-by-step explanation:

The probability that the birth is a male can be written as

$p(m) = 0.52$ (which corresponds to 52%)

While the probability that the birth is a female can be written as

$p(f) = 0.48$ (which corresponds to 48%)

Here we want to calculate the probability that over 2 births, both are male. Since the two births are two independent events (the probability of the 2nd to be a male does not depend on the fact that the 1st one is a male), then the probability of both being males is given by the product of the individual probabilities:

$p(mm)=p(m)\cdot p(m)$

And substituting, we find

$p(mm)=0.52\cdot 0.52 = 0.27$

So, 27%.

In this case, we want to find the probability that both children are female, so the probability

$p(ff)$

As in the previous case, the probability of the 2nd child to be a female is independent from whether the 1st one is a male or a female: therefore, we can apply the rule for independent events, and this means that the probability that both children are females is the product of the individual probability of a child being a female:

$p(ff)=p(f)\cdot p(f)$