The Prime Factorization of 760 is 19·5·2·2·2
Because those are the prime number factors of 760
Answer:
8
Step-by-step explanation:
we can say that AD is congruent to DC
so your equation for x is: 4x - 1 = 2x
solve this to get x = 0.5
plug x into and equation and multiply you answer by 2 to find the hypotenuse of triangle ABC and DEF
4(0.5) - 1 = 1
hypotenuse: 1 x 2 = 2
since we know x is 0.5, plug this into 4x + 1 to find the length of the leg FE,
4(0.5) + 1 = 3
In the diagram, it shows that the legs of triangle are congruent
this means that FE, ED, BA, and BC are all congruent
since we know FE is 3, we know that all the other sides are 3 as well
this means that the perimeter of the triangle is: leg + leg + hypotenuse
so 3 + 3 + 2
the perimeter is 8
Answer:
B. 4200
Step-by-step explanation:
A suitable calculator can add the terms for you. (See attached)
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The first term is 69, and the common difference is 75-69 = 6. The general term is ...
an = a1 +d(n -1)
an = 69 +6(n -1)
Then the 28th term is ...
a28 = 69 +6(27) = 231
The average term is (a28 +a1)/2 = (231 +69)/2 = 150.
The sum is the number of terms multiplied by the average term:
sum = 28×150= 4200
Answer:
A)16
Step-by-step explanation:
Given
f(x)=5x^2+9x-2
Remainder theorem states that when f(x) is divided by x-a then the remainder can be calculated by calculating f(a).
Now Using the remainder theorem to divide 5x^2+9x-2 by x+3 to find the remainder:
f(x)=5x^2+9x-2
f(-3) = 5(-3)^2 +9(-3) -2
=5(9) - 27 -2
= 45-29
= 16 !
Answer:
f'(1) = 2
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Algebra I</u>
<u>Calculus</u>
The definition of a derivative is the slope of the tangent line.
Derivative Notation
Basic Power Rule:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Step-by-step explanation:
<u>Step 1: Define</u>
f(x) = x²
Point (1, f(1))
<u>Step 2: Differentiate</u>
- Basic Power Rule: f'(x) = 2 · x²⁻¹
- Simplify: f'(x) = 2x
<u>Step 3: Find Slope</u>
<em>Use the point (1, f(1)) to find the instantaneous slope</em>
- Substitute in <em>x</em>: f'(1) = 2(1)
- Multiply: f'(1) = 2
This tells us that at point (1, f(1)), the slope of the tangent line is 2. We can write an equation using point slope form as well: y - f(1) = 2(x - 1)
