Match each expression with an equivalent expression.

Kara had a savings account balance of $153 on Monday on Tuesday she has six withdrawals of $15.72 and a deposit of $235.15 what

trasher [3.6K]

Answer:

$293.83

Step-by-step explanation:

She started with $153. The next day she withdrew $15.72 six times, that is $15.72*6 = $94.32

We have to subtract $94.32 from the starting amount, that is $153 - $94.32 = $58.68

After that, she deposited $235.15. We have to add that to the current balance.

Now she has $58.68 + $235.15 = $293.83

Therefore, her account balance after these transactions is $293.83

3 0

3 years ago

Two angles are vertical. One angle is by degrees and the other angle

Liono4ka [1.6K]

Answer:

Supplementary angles are two angles whose measures add up to 180 degrees. ... Supplementary angles are two angles whose measures add up to 180° . The two angles of a linear pair , like ∠1 and ∠2 in the figure below, are ... If the measure of the angle is twice the measure of the other, find the measure of each angle.

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

HURRY!!!! NO SPAM!!!! Identify the area of the trapezoid.

saveliy_v [14]

Answer:

$area = >$

a+b/2*h

Here,

a = 12xcm

b= 15xcm

h= 7cm

$= >$

(12+15)xcm/2*7

=27xcm*3.5

8 0

3 years ago

Read 2 more answers

The fact that the polynomial (3x^2+4x+5)+(2x+6) Fields a polynomial employs what property?

EleoNora [17]

Answer:

The fact that the polynomial (3x2+4x+5) + (2x+6) yields a poloynomial, employs what property?

Step-by-step explanation:

3 0

3 years ago

NO LINKS!!! What is the equation of these two graphs?

S_A_V [24]

Answer:

$\textsf{1.} \quad y=-\dfrac{1}{30}x^2+\dfrac{1}{2}x+\dfrac{68}{15}\:\:\textsf{ where}\:\:x \geq \dfrac{15-\sqrt{769}}{2}\\\\\quad \textsf{or} \quad y=2\sqrt{x+5}$

$\textsf{2.} \quad y=-|x+1|+5$

Step-by-step explanation:

<h3><u>Question 1</u></h3>

<u>Method 1 - modelling as a quadratic with restricted domain</u>

Assuming that the points given on the graph are points that the <u>curve passes through</u>, the curve can be modeled as a quadratic with a limited domain. Please note that as the x-intercept has not been defined on the graph, I am not including this in this first method.

Standard form of a quadratic equation:

$y=ax^2+bx+c$

Given points:

(-4, 2)
(-1, 4)
(4, 6)

Substitute the given points into the equation to create 3 equations:

<u>Equation 1 (-4, 2)</u>

$\implies a(-4)^2+b(-4)+c=2$

$\implies 16a-4b+c=2$

<u>Equation 2 (-1, 4)</u>

$\implies a(-1)^2+b(-1)+c=4$

$\implies a-b+c=4$

<u>Equation 3 (4, 6)</u>

$\implies a(4)^2+b(4)+c=6$

$\implies 16a+4b+c=6$

Subtract Equation 1 from Equation 3 to eliminate variables a and c:

$\implies (16a+4b+c)-(16a-4b+c)=6-2$

$\implies 8b=4$

$\implies b=\dfrac{4}{8}$

$\implies b=\dfrac{1}{2}$

Subtract Equation 2 from Equation 3 to eliminate variable c:

$\implies (16a+4b+c)-(a-b+c)=6-4$

$\implies 15a+5b=2$

$\implies 15a=2-5b$

$\implies a=\dfrac{2-5b}{15}$

Substitute found value of b into the expression for a and solve for a:

$\implies a=\dfrac{2-5(\frac{1}{2})}{15}$

$\implies a=-\dfrac{1}{30}$

Substitute found values of a and b into Equation 2 and solve for c:

$\implies a-b+c=4$

$\implies -\dfrac{1}{30}-\dfrac{1}{2}+c=4$

$\implies c=\dfrac{68}{15}$

Therefore, the equation of the graph is:

$y=-\dfrac{1}{30}x^2+\dfrac{1}{2}x+\dfrac{68}{15}$

$\textsf{with the restricted domain}: \quad x \geq \dfrac{15-\sqrt{769}}{2}$

<u>Method 2 - modelling as a square root function</u>

Assuming that the points given on the graph are points that the <u>curve passes through</u>, and the x-intercept should be included, we can model this curve as a <u>square root function</u>.

Given points: