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pav-90 [236]
2 years ago
11

Tara had $350 in her bank account she received three birthday checks for $35 each and but bought two new video games for $41 eac

h
Mathematics
1 answer:
goldfiish [28.3K]2 years ago
6 0

We conclude that the final amount that she has in her bank account is $375.

<h3>How much has Tara in her bank account now?</h3>

Here we need to perform some additions and subtractions. We know that Tara starts with $350 in her account.

Then she received 3 checks for $35 each, so at this point she has:

$350 + 3*$35 = $455

Now she buys 2 video games for $41 each, so we need to subtract two times $41, we will get:

$455 - 2*$41 = $375

We conclude that the final amount that she has in her bank account is $375.

If you want to learn more about additions and subtractions:

brainly.com/question/25421984

#SPJ1

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Figure ABCD is a kite. The area of ABCD is 48 square units. The length of line segment BD is 8 units. What is the length of AC?
astra-53 [7]
The area of a kite:
A = d1 * d2 / 2
d1 = BD = 8 units
48 = 8 * AC / 2
48 * 2 = 8 AC
96 = 8 AC
AC = 96 : 8 = 12
Answer: D ) 12 units
7 0
3 years ago
What is the derivative of y with respect to x if y = 2/X4
zepelin [54]
You're correct, the answer is C.

Given any function of the form y=x^n, then the derivative of y with respect to x (\frac{dy}{dx}) is written as:
\frac{dy}{dx}=nx^n^-^1

In which n is any constant, this is called the power rule for differentiation.

For this example we have y= \frac{2}{x^4}, first lets get rid of the quotient and write the expression in the form y=x^n:
y= \frac{2}{x^4} =2x^-^4

Now we can directly apply the rule stated at the beginning (in which n=-4):
\frac{dy}{dx}=(2)(-4)x^{-4-1}=-8 x^{-5}= -\frac{8}{x^5}

Note that whenever we differentiate a function, we simply "ignore" the constants (we take them out of the derivative).
6 0
3 years ago
Evaluate the given integral by changing to polar coordinates. 8xy dA D , where D is the disk with center the origin and radius 9
BabaBlast [244]

Answer:

0

Step-by-step explanation:

∫∫8xydA

converting to polar coordinates, x = rcosθ and y = rsinθ and dA = rdrdθ.

So,

∫∫8xydA = ∫∫8(rcosθ)(rsinθ)rdrdθ = ∫∫8r²(cosθsinθ)rdrdθ = ∫∫8r³(cosθsinθ)drdθ

So we integrate r from 0 to 9 and θ from 0 to 2π.

∫∫8r³(cosθsinθ)drdθ = 8∫[∫r³dr](cosθsinθ)dθ

= 8∫[r⁴/4]₀⁹(cosθsinθ)dθ

= 8∫[9⁴/4 - 0⁴/4](cosθsinθ)dθ

= 8[6561/4]∫(cosθsinθ)dθ

= 13122∫(cosθsinθ)dθ

Since sin2θ = 2sinθcosθ, sinθcosθ = (sin2θ)/2

Substituting this we have

13122∫(cosθsinθ)dθ = 13122∫(1/2)(sin2θ)dθ

= 13122/2[-cos2θ]/2 from 0 to 2π

13122/2[-cos2θ]/2 = 13122/4[-cos2(2π) - cos2(0)]

= -13122/4[cos4π - cos(0)]

= -13122/4[1 - 1]

= -13122/4 × 0

= 0

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