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2 years ago

I need help please can u please fill out AA aa Aa please i will give braneist

Mathematics

1 answer:

maks197457 [2]2 years ago

7 0

Hey!

Your answers are:

<h3>AA: Normal (Unaffected)</h3><h3>Aa: Normal (Carrier)</h3><h3>aa: Albino (Effected)</h3>

<em>Hope this helps!</em>

<h2><em>~~~PicklePoppers~~~</em></h2>

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Find an approximate equation y=ab^x with the coordinates (0, 4) (3, 95)

bogdanovich [222]

We have $y=ab^{x}$

Plug in $(0, 4)$ :
$4=ab^{0}$ ⇒ $4=a$
So we now have $a$

Plug in $(3, 95)$ :
$95=4b^{3}$
⇒ $b^{3}=\frac{95}{4}$
⇒ $b=\sqrt[3]{\frac{95}{4}}$
which is approximately 2.874

So we get
$y=4(\sqrt[3]{\frac{95}{4}})^{x}$
or, in decimal form,
$y=4*2.874^{x}$

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3 years ago

Jason’s savings account has a balance of $2179. After 5 years , what will the amount of interest be at 6% compounded quarterly?

satela [25.4K]

Answer:

$755.80

Step-by-step explanation:

Determine the compound amount first and then subtract the principal from it, to find the amount of interest.

The compound amount formula is A = P (1 + r/n)^(nt), where

P is the initial principal, r is the interest rate as a decimal fraction, n is the number of compounding periods per year, and t is the number of years. Here, P = $2179; t = 5 yrs; r = 0.06; and n = 4 (quarterly compounding).

We get:

A = $2179(1 + 0.06/4)^(4*5), or $2179(1.015)^20, or $2179(1.347) = $2937.80.

The compound amount is $2934.80. Subtracting the $2179 principal results in the interest earned: $755.80.

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3 years ago

Find a general solution of y" + 4y = 0.

klio [65]

Answer:

$y(x)=C_{1}cos2x+C_{2}sin(2x)$

Step-by-step explanation:

It is a linear homogeneous differential equation with constant coefficients:

y" + 4y = 0

Its characteristic equation:

r^2+4=0

r1=2i

r2=-2i

We use these roots in order to find the general solution:

$y(x)=C_{1}cos2x+C_{2}sin(2x)$

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3 years ago

Given: △ACM, m∠C=90°, CP ⊥ AM

larisa86 [58]

Answer: The answer is $3\dfrac{4}{7}.$

Step-by-step explanation: Given in the question that ΔAM is a right-angled triangle, where ∠C = 90°, CP ⊥ AM, AC : CM = 3 : 4 and MP - AP = 1. We are to find AM.

Let, AC = 3x and CM = 4x.

In the right-angled triangle ACM, we have

$AM^2=AC^2+CM^2=(3x)^2+(4x)^2=9x^2+16x^2=25x^2\\\\\Rightarrow AM=5x.$

Now,

$AM=AP+PM=AP+(AP+1)\\\\\Rightarrow 2AP=AM-1\\\\\Rightarrow 2AP=5x-1.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(A)$

Now, since CP ⊥ AM, so ΔACP and ΔMCP are both right-angled triangles.

So,

$CP^2=AC^2-AP^2=CM^2-MP^2\\\\\Rightarrow (3x)^2-AP^2=(4x)^2-(AP+1)^2\\\\\Rightarrow 9x^2-AP^2=16x^2-AP^2-2AP-1\\\\\Rightarrow 2AP=7x^2-1.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(B)$