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3 years ago

What is the blood type of the unknown parent?

Mathematics

1 answer:

Vinil7 [7]3 years ago

5 0

Answer:

IBi0

Step-by-step explanation:

We look at the child's genotype (lAi0).

We know that the blood type's genotype has 2 genes, one from the mother and one from the father.

If the genotype is lAi0 and one gene is, let's say, from mother (lA), we know exactly that the other gene is from father (i0).

So, father's genotype is lBi0 (a).

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Please answer this correctly

motikmotik

I would say answer 3

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3 years ago

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The area of the rectangular faces of the box are shown . What is the volume of the box

Gnoma [55]

Label the 3 distinct sides of the box. I arbitrarily chose the letters a, b and c.
Use the info about areas as follows:

ab=54 in^2
ac=90 in^2
bc=60 in^2

Here you have 3 equations in 3 unknowns (a, b and c), which is enough info to use to determine a, b and c. Then the volume of the box is a*b*c.

Example: bc = 90, but c = 60/b. You could subst. 60/b for c in the 2nd and 3rd equation, which will eliminate c completely and leave you with 2 equations in 2 unknowns.

Continuing this procedure, I determined that a=9, b=6 and c=10. Thus, the volume of the box is V = 9*6*10 = 540 cubic inches (answer)

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3 years ago

Mike owes his brother $9.55. He borrows $10.75 more from his brother. Which of the following best represents Mike's debt to his

Radda [10]

Answer:

C - $20.30

Step-by-step explanation:

-9.55 + (-10.75) = - 20.3

8 0

3 years ago

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Please dont ignore, Need help!!! Use the law of sines/cosines to find..

Ket [755]

Answer:

16. Angle C is approximately 13.0 degrees.

17. The length of segment BC is approximately 45.0.

18. Angle B is approximately 26.0 degrees.

15. The length of segment DF "e" is approximately 12.9.

Step-by-step explanation:

By the law of sine, the sine of interior angles of a triangle are proportional to the length of the side opposite to that angle.

For triangle ABC:

$\sin{A} = \sin{103\textdegree{}}$ ,
The opposite side of angle A $a = BC = 26$ ,
The angle C is to be found, and
The length of the side opposite to angle C $c = AB = 6$ .

$\displaystyle \frac{\sin{C}}{\sin{A}} = \frac{c}{a}$ .

$\displaystyle \sin{C} = \frac{c}{a}\cdot \sin{A} = \frac{6}{26}\times \sin{103\textdegree}$ .

$\displaystyle C = \sin^{-1}{(\sin{C}}) = \sin^{-1}{\left(\frac{c}{a}\cdot \sin{A}\right)} = \sin^{-1}{\left(\frac{6}{26}\times \sin{103\textdegree}}\right)} = 13.0\textdegree{}$ .

Note that the inverse sine function here $\sin^{-1}()$ is also known as arcsin.

By the law of cosine,

$c^{2} = a^{2} + b^{2} - 2\;a\cdot b\cdot \cos{C}$ ,

where

$a$ , $b$ , and $c$ are the lengths of sides of triangle ABC, and
$\cos{C}$ is the cosine of angle C.

For triangle ABC:

$b = 21$ ,
$c = 30$ ,
The length of $a$ (segment BC) is to be found, and
The cosine of angle A is $\cos{123\textdegree}$ .

Therefore, replace C in the equation with A, and the law of cosine will become:

$a^{2} = b^{2} + c^{2} - 2\;b\cdot c\cdot \cos{A}$ .

$\displaystyle \begin{aligned}a &= \sqrt{b^{2} + c^{2} - 2\;b\cdot c\cdot \cos{A}}\\&=\sqrt{21^{2} + 30^{2} - 2\times 21\times 30 \times \cos{123\textdegree}}\\&=45.0 \end{aligned}$ .

For triangle ABC:

$a = 14$ ,
$b = 9$ ,
$c = 6$ , and
Angle B is to be found.

Start by finding the cosine of angle B. Apply the law of cosine.

$b^{2} = a^{2} + c^{2} - 2\;a\cdot c\cdot \cos{B}$ .

$\displaystyle \cos{B} = \frac{a^{2} + c^{2} - b^{2}}{2\;a\cdot c}$ .

$\displaystyle B = \cos^{-1}{\left(\frac{a^{2} + c^{2} - b^{2}}{2\;a\cdot c}\right)} = \cos^{-1}{\left(\frac{14^{2} + 6^{2} - 9^{2}}{2\times 14\times 6}\right)} = 26.0\textdegree$ .