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2 years ago

What is the sum (-19X10) + 7

Mathematics

1 answer:

Scorpion4ik [409]2 years ago

6 0

Answer:

-19×10= -190

-190+7= -183

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H(t) = -16t^2 + 40t +120

stira [4]

Answer:

h(t) = -16t^2 + 40t + 120

h(t) = -16(t^2 - 5/2*t) + 120

h(t) = -16(t^2 - 5/2*t + (5/4)^2 - (5/4)^2) + 120

h(t) = -16(t^2 - 5/2*t + (5/4)^2 - 25/16) + 120

h(t) = -16(t^2 - 5/2*t + (5/4)^2) + 120 + 25

h(t) = -16(t - 5/4)^2 + 145

The rock needs 5/4 = 1.25 seconds to travel to the highest point wich is 145 m high.

6 0

3 years ago

Reduce then calculate 7/8 x 4/15 x 3/7 = ? Can you help?

Zepler [3.9K]

since this is a multiplication, all the numerators are just factors of the product numerator and all denominators are just factors of the product denominator, so we can simply reorder them some, without changing the product.

$\bf \cfrac{7}{8}\times \cfrac{4}{15}\times \cfrac{3}{7}\implies \cfrac{7}{7}\times \cfrac{4}{8}\times \cfrac{3}{15}\implies \cfrac{1}{1}\times \cfrac{1}{2}\times \cfrac{1}{5}\implies \cfrac{1\cdot 1\cdot 1}{1\cdot 2\cdot 5}\implies \cfrac{1}{10}$

8 0

2 years ago

(05.01)

choli [55]

Step-by-step explanation:

The graph of a system of parallel lines will have no solutions.

Always

Because they will never intersect, having same slopes.

=========================

The graph of a system of equations with different slopes will have no solutions.

Never

Because they will intersect exactly once, having different slopes.

6 0

1 year ago

Read 2 more answers

Suppose that the length of a side of a cube X is uniformly distributed in the interval 9

Nastasia [14]

Answer:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

Step-by-step explanation:

Given

$9 < x < 10$ --- interval

Required

The probability density of the volume of the cube

The volume of a cube is:

$v = x^3$

For a uniform distribution, we have:

$x \to U(a,b)$

and

$f(x) = \left \{ {{\frac{1}{b-a}\ a \le x \le b} \atop {0\ elsewhere}} \right.$

$9 < x < 10$ implies that:

$(a,b) = (9,10)$

So, we have:

$f(x) = \left \{ {{\frac{1}{10-9}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

Solve

$f(x) = \left \{ {{\frac{1}{1}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

$f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

Recall that:

$v = x^3$

Make x the subject

$x = v^\frac{1}{3}$

So, the cumulative density is:

$F(x) = P(x < v^\frac{1}{3})$

$f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$ becomes

$f(x) = \left \{ {{1\ 9 \le x \le v^\frac{1}{3} - 9} \atop {0\ elsewhere}} \right.$

The CDF is:

$F(x) = \int\limits^{v^\frac{1}{3}}_9 1\ dx$

Integrate

$F(x) = [v]\limits^{v^\frac{1}{3}}_9$

Expand

$F(x) = v^\frac{1}{3} - 9$

The density function of the volume F(v) is:

$F(v) = F'(x)$

Differentiate F(x) to give:

$F(x) = v^\frac{1}{3} - 9$

$F'(x) = \frac{1}{3}v^{\frac{1}{3}-1}$

$F'(x) = \frac{1}{3}v^{-\frac{2}{3}}$

$F(v) = \frac{1}{3}v^{-\frac{2}{3}}$

So:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

8 0

2 years ago

Lorne subtracted 6x3 – 2x + 3 from –3x3 + 5x2 + 4x – 7. Use the drop-down menus to identify the steps Lorne used to find the dif

Novay_Z [31]

Answer: Step 1: Reverse the signs of $6x^3-2x + 3$ expression.

Step 2: Removing parenthesis

Step 3: Grouping like terms

Step 4: Combing like terms

Step 5: Writing the final expression in standard form

Step-by-step explanation: First expression : $6x^3-2x + 3$

Second expression : $-3x^3+5x^2+4x-7$ .

We need to subtract $6x^3-2x + 3$ from $-3x^3+5x^2+4x-7$ .

Step 1: Reverse the signs of $6x^3-2x + 3$ expression.

$(-3x^3+5x^2+4x-7)+(-6x3+2x-3)$

Step 2: Removing parenthesis

$(-3x^3) +5x^2+4x + (-7) + (-6x^3) + 2x + (-3)$

Step 3: Grouping like terms

$[-3x^3) + (-6x^3)] + [4x + 2x] + [(-7) + (-3)] + [5x^2]$

Step 4: Combing like terms

$-9x^3 + 6x + (-10) + 5x^2$

Step 5: Writing the final expression in standard form

$-9x^3 + 5x^2 + 6x-10$

8 0

2 years ago

Read 2 more answers