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sattari [20]
2 years ago
15

A=bx+cx/2 please help

Mathematics
1 answer:
andreyandreev [35.5K]2 years ago
5 0

Answer:

I'm too tired to explain, here's some digital work on a site called Symbolab:

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Aubrey’s monthly bank statement shows a total of $51 in fees for ATM withdrawals. If Audrey made 17 withdraws, what is Audrey ch
Tpy6a [65]

Answer:

Audrey was charged $3 for each withdrawal.

Step-by-step explanation:

We have to calculate the amount of money charged on one transaction.

In order to find the amount for one ATM withdrawal, the total amount for ATM withdrawals will be divided by number of withdrawals.

Given

Total\ ATM\ withdrawal\ fees = \$51\\Total\ withdrawals = 17\\Amount\ for\ one\ withdrawal = \frac{Total\ fees}{No.\ of\ withdrawals}\\= \frac{51}{17}\\= \$3

Hence,

Audrey was charged $3 for each withdrawal.

7 0
3 years ago
A fruit vendor sells 85 pieces of fruit that are either apples or oranges. The ratio of apples to oranges is 3:2. How many apple
crimeas [40]
53. The ratio of apple over orange = 3:2 The total number of fruits are 60 pieces. Now, find the total number of apples and total number of oranges. => first, let’s add 2 + 3 = 5 Thus, we have 5 division for 60 pieces fruits, => 60 / 5 => 12 let’s solve for the apple => 12 x 3 = 36 Now. The orange => 12 * 2 = 24 => 36 + 24 = 60 Thus the ratio is 36 : 24
5 0
3 years ago
Solve for x. Enter the solutions from least to greatest.
Len [333]

Answer:

This is the answer of your question ☺☺

Step-by-step explanation:

(x+7)2 -49=0

2x+14-49=0

2x-35=0

2x=35

x=17.5

7 0
2 years ago
the vertex form of the equation of a parabola is y=5(x-3)2-6. What is the standard form of the equation
kobusy [5.1K]
The answer is
y=5x^2-30x+39
(apex)
5 0
3 years ago
Read 2 more answers
(sinA + cosA)/ (secA + cosecA) = sinA* cosA​
OlgaM077 [116]

Answer:

<em>Proof below</em>

Step-by-step explanation:

<u>Trigonometric Identities</u>

We'll prove that:

\displaystyle \frac{\sin A+\cos A}{\sec A+\csc A}=\sin A*\cos A

Recall:

\displaystyle \sec A =\frac{1}{\cos A}

\displaystyle \csc A =\frac{1}{\sin A}

Applying those definitions:

\displaystyle \frac{\sin A+\cos A}{\sec A+\csc A}=\frac{\sin A+\cos A}{\frac{1}{\cos A}+\frac{1}{\sin A}}

Adding the fractions:

\displaystyle \frac{\sin A+\cos A}{\sec A+\csc A}=\frac{\sin A+\cos A}{\frac{\sin A+\cos A}{\cos A*\sin A}}

Dividing the fractions:

\displaystyle \frac{\sin A+\cos A}{\sec A+\csc A}=(\sin A+\cos A)*\frac{\cos A*\sin A}{\sin A+\cos A}

Simplifying:

\displaystyle \frac{\sin A+\cos A}{\sec A+\csc A}=\cos A*\sin A

Hence proved

6 0
3 years ago
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