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4 years ago

3. Solve the inequality –2(z + 5) + 20 > 6.

Mathematics

1 answer:

12345 [234]4 years ago

4 0

$\boxed{-2\left(z+5\right)+20>6\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:z$

Solution:

Given inequality is:

$-2(z+5)+20>6$

We have to solve the given inequality

$-2\left(z+5\right)+20>6\\\\\mathrm{Subtract\:}20\mathrm{\:from\:both\:sides}\\\\-2\left(z+5\right)+20-20>6-20\\\\\mathrm{Simplify}\\\\-2\left(z+5\right)>-14$

$Multiply\ both\ sides\ by\ -1\ \left(reverse\:the\:inequality\right)$

Whenever we multiply or divide an inequality by a negative number, we must flip the inequality sign

$\left(-2\left(z+5\right)\right)\left(-1\right)$

Thus the solution to inequality is:

$-2\left(z+5\right)+20>6\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:z$

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18.96 x 2.03 correct to 2 significant figures equals what?

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Answer:

Step-by-step explanation:

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3 years ago

Orthogonalizing vectors. Suppose that a and b are any n-vectors. Show that we can always find a scalar γ so that (a − γb) ⊥ b, a

jeka57 [31]

Answer:

$\\ \gamma= \frac{a\cdot b}{b\cdot b}$

Step-by-step explanation:

The question to be solved is the following :

Suppose that a and b are any n-vectors. Show that we can always find a scalar γ so that (a − γb) ⊥ b, and that γ is unique if $b \neq 0$ . Recall that given two vectors a,b a⊥ b if and only if $a\cdot b =0$ where $\cdot$ is the dot product defined in $\mathbb{R}^n$ . Suposse that $b\neq 0$ . We want to find γ such that $(a-\gamma b)\cdot b=0$ . Given that the dot product can be distributed and that it is linear, the following equation is obtained

$(a-\gamma b)\cdot b = 0 = a\cdot b - (\gamma b)\cdot b= a\cdot b - \gamma b\cdot b$

Recall that $a\cdot b, b\cdot b$ are both real numbers, so by solving the value of γ, we get that

$\gamma= \frac{a\cdot b}{b\cdot b}$

By construction, this γ is unique if $b\neq 0$ , since if there was a $\gamma_2$ such that $(a-\gamma_2b)\cdot b = 0$ , then

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4 years ago

Find the distance between the points G(-5, 4) and H(2, 6).

vladimir2022 [97]

Answer:

The answer is

<h2> $\sqrt{53} \: \: or \: \: 7.28 \: \: \: \: units$ </h2>

Step-by-step explanation:

The distance between two points can be found by using the formula

$d = \sqrt{ ({x1 - x2})^{2} + ({y1 - y2})^{2} } \\$

where

(x1 , y1) and (x2 , y2) are the points

From the question the points are

G(-5, 4) and H(2, 6)

The distance between them is

$|GH | = \sqrt{ ({ - 5 - 2})^{2} + ({4 - 6})^{2} } \\ = \sqrt{ ({ - 7})^{2} + ( { - 2})^{2} } \: \\ = \sqrt{49 + 4} \\ = \sqrt{53} \: \: \: \: \: \: \: \: \\ = 7.280109$