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3 years ago

Please helpppp I will give you a brainleist !!!!!!

Mathematics

2 answers:

Verdich [7]3 years ago

4 0

Answer:

Hope this helps

Step-by-step explanation:

4. 10^2

5. 64 / 14 * 9 + 8

64/126+8

8 64/126

8 4/9

6. 1.762 x 10^3

Lyrx [107]3 years ago

3 0

Answer:

I am confused with #5 but the rest are solid and checked

Step-by-step explanation:

#4 (A)

10*10=100

#5 (97)

64+4*4+3*3+2*2*2=

64+16+9+8=97

#6 (1.762 × 103)

#7 1070

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If f(x) =4x2 - 8x - 20 and g(x) = 2x + a, find the value of a so that the y-intercept of the graph of the composite function (fo

amm1812

Answer:

The possible values are a = -2.5 or a = 4.5.

Step-by-step explanation:

Composite function:

The composite function of f(x) and g(x) is given by:

$(f \circ g)(x) = f(g(x))$

In this case:

$f(x) = 4x^2 - 8x - 20$

$g(x) = 2x + a$

$(f \circ g)(x) = f(g(x)) = f(2x + a) = 4(2x + a)^2 - 8(2x + a) - 20 = 4(4x^2 + 4ax + a^2) - 16x - 8a - 20 = 16x^2 + 16ax + 4a^2 - 16x - 8a - 20 = 16x^2 +(16a-16)x + 4a^2 - 8a - 20$

Value of a so that the y-intercept of the graph of the composite function (fog)(x) is (0, 25).

This means that when $x = 0, f(g(x)) = 25$ . So

$4a^2 - 8a - 20 = 25$

$4a^2 - 8a - 45 = 0$

Solving a quadratic equation, by Bhaskara:

$\Delta = (-8)^2 - 4(4)(-45) = 784$

$x_{1} = \frac{-(-8) + \sqrt{784}}{2*(4)} = \frac{36}{8} = 4.5$

$x_{2} = \frac{-(-8) - \sqrt{784}}{2*(4)} = -\frac{20}{8} = -2.5$

The possible values are a = -2.5 or a = 4.5.

5 0

3 years ago

A = 1011 + 337 + 337/2 +1011/10 + 337/5 + ... + 1/2021

egoroff_w [7]

The sum of the given series can be found by simplification of the number

of terms in the series.

A is approximately <u>2020.022</u>

Reasons:

The given sequence is presented as follows;

A = 1011 + 337 + 337/2 + 1011/10 + 337/5 + ... + 1/2021

Therefore;

$\displaystyle A = \mathbf{1011 + \frac{1011}{3} + \frac{1011}{6} + \frac{1011}{10} + \frac{1011}{15} + ...+\frac{1}{2021}}$

The n + 1 th term of the sequence, 1, 3, 6, 10, 15, ..., 2021 is given as follows;

$\displaystyle a_{n+1} = \mathbf{\frac{n^2 + 3 \cdot n + 2}{2}}$

Therefore, for the last term we have;

$\displaystyle 2043231= \frac{n^2 + 3 \cdot n + 2}{2}$

2 × 2043231 = n² + 3·n + 2

Which gives;

n² + 3·n + 2 - 2 × 2043231 = n² + 3·n - 4086460 = 0

Which gives, the number of terms, n = 2020

$\displaystyle \frac{A}{2} = \mathbf{ 1011 \cdot \left(\frac{1}{2} +\frac{1}{6} + \frac{1}{12}+...+\frac{1}{4086460} \right)}$

$\displaystyle \frac{A}{2} = 1011 \cdot \left(1 - \frac{1}{2} +\frac{1}{2} - \frac{1}{3} + \frac{1}{3}- \frac{1}{4} +...+\frac{1}{2021}-\frac{1}{2022} \right)$