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Anestetic [448]
3 years ago
8

Kurt deposited money into his bank account every month. The amount of money he deposited in the first 5 months of 2014 is listed

in the table.
Month Amount Deposited
(dollars)
January 1,200.32
February 985.43
March 1,200.65
April 1,987.34
May 1,000.98
The total amount of money that Kurt had deposited into his account at the end of May is ______
The total amount of money that he had deposited into his account at the end of March is _______
Mathematics
2 answers:
Irina-Kira [14]3 years ago
3 1

Answer:

that is the answer

Step-by-step explanation:

zimovet [89]3 years ago
4 0
You have the answer I’m pretty sure
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Find KL.<br> K<br> X-1<br> L<br> 12<br> M
Aneli [31]

Answer:

KL = 9

Step-by-step explanation:

(7 + x - 1)(x - 1) = 12²

(x + 6)(x - 1) = 144

x² -x + 6x - 6 = 144

reduce:

x² + 5x -150 = 0

(x + 15)(x - 10) = 0

x = -15, x = 10

KL = 10 - 1 = 9

6 0
3 years ago
Divide the rational expressions and express in simplest form. When typing your answer for the numerator and denominator be sure
Veseljchak [2.6K]

Dividing by a fraction is equivalent to multiply by its reciprocal, then:

\begin{gathered} \frac{3y^2-7y-6}{2y^2-3y-9}\div\frac{y^2+y-2}{2y^2+y-3^{}}= \\ =\frac{3y^2-7y-6}{2y^2-3y-9}\cdot\frac{2y^2+y-3}{y^2+y-2}= \\ =\frac{(3y^2-7y-6)(2y^2+y-3)}{(2y^2-3y-9)(y^2+y-2)} \end{gathered}

Now, we need to express the quadratic polynomials using their roots, as follows:

ay^2+by+c=a(y-y_1)(y-y_2)

where y1 and y2 are the roots.

Applying the quadratic formula to the first polynomial:

\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{7\pm\sqrt[]{(-7)^2-4\cdot3\cdot(-6)}}{2\cdot3} \\ y_{1,2}=\frac{7\pm\sqrt[]{121}}{6} \\ y_1=\frac{7+11}{6}=3 \\ y_2=\frac{7-11}{6}=-\frac{2}{3} \end{gathered}

Applying the quadratic formula to the second polynomial:

\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot2\cdot(-3)}}{2\cdot2} \\ y_{1,2}=\frac{-1\pm\sqrt[]{25}}{4} \\ y_1=\frac{-1+5}{4}=1 \\ y_2=\frac{-1-5}{4}=-\frac{3}{2} \end{gathered}

Applying the quadratic formula to the third polynomial:

\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{3\pm\sqrt[]{(-3)^2-4\cdot2\cdot(-9)}}{2\cdot2} \\ y_{1,2}=\frac{3\pm\sqrt[]{81}}{4} \\ y_1=\frac{3+9}{4}=3 \\ y_2=\frac{3-9}{4}=-\frac{3}{2} \end{gathered}

Applying the quadratic formula to the fourth polynomial:

\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot1\cdot(-2)}}{2\cdot1} \\ y_{1,2}=\frac{-1\pm\sqrt[]{9}}{2} \\ y_1=\frac{-1+3}{2}=1 \\ y_2=\frac{-1-3}{2}=-2 \end{gathered}

Substituting into the rational expression and simplifying:

\begin{gathered} \frac{3(y-3)(y+\frac{2}{3})2(y-1)(y+\frac{3}{2})}{2(y-3)(y+\frac{3}{2})(y-1)(y+2)}= \\ =\frac{3(y+\frac{2}{3})}{2(y+2)}= \\ =\frac{3y+2}{2y+4} \end{gathered}

8 0
1 year ago
What is the sum of the complex number 2+3i and 4+8i, where i=√-1?
ser-zykov [4K]
2+3i+4+8i
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6+11i
3 0
3 years ago
Jamal draws triangle XYZ. If angle X=46 degrees, XZ = 10 degrees, and YZ=8 degrees, what is the approximate length of XY?
Mashutka [201]
<span>In triangle XYZ,
angle X=46 degrees,
XZ = 10 units and
YZ=8 units.

We will find the length of XY using cosine rule:

According to cosine rule:

a² = b² + c² - 2bc cosA (the figure is attached)
Here, A = 46 degree.
a = 8 units.
b <span>= ?
</span>c = 10 units.
</span>⇒  b² = c² - a² -2bc cosA
⇒  b² = 100 - 64 - 160*cos(46)
⇒  b² = 100 - 64 - 160*0.69
⇒  b² = 100 - 64 - 110.4
⇒  b² = 100 - 64 - 110.4
⇒ b² = 74.4
or b = 8.62 units.
or b = 8 untis approximately. 

5 0
3 years ago
Read 2 more answers
Alex has $25 in his savings account. He needs $150 to buy a new bicycle. He plans to deposit $12.50 per week into the account un
Mice21 [21]

Answer:

150-25=125 (what he needs left)

125/12.50= 10

So it'll take him 10 weeks to finally reach that amount.

Step-by-step explanation:

4 0
3 years ago
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