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2 years ago

What is the value of the expression

Mathematics

2 answers:

sleet_krkn [62]2 years ago

7 0

Answer:

the value of expression is 1\2

EleoNora [17]2 years ago

4 0

Sjshsbsbsbsnsnsjdnnd

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14 5 18 24 15 28 34 _ what number is needed to complete the pattern

AlekseyPX

Minus 9, add 13, add 6, mus 9, add 13, add 6, minus 9 is next

34-9=25
the blank is 25

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3 years ago

In △ABC, ∠ABC = 60○

Temka [501]

A given shape that is bounded by three sides and has got three internal angles is referred to as a triangle. Thus the value of PB is 8.0 units.

A given shape that is bounded by three sides and has got three internal angles is referred to as a triangle. Types of triangles include right angle triangle, isosceles triangle, equilateral triangle, acute angle triangle, etc. The sum of the internal angles of any triangle is $180^{o}$ .

In the given question, point P is such that <APB = <APC = <BPC = $120^{o}$ . Also, line PB bisects <ABC into two equal measures. Thus;

<ABP = $30^{o}$

Thus,

<ABP + <APB + <BAP = $180^{o}$

30 + 120 + <BAP = $180^{o}$

<BAP = $180^{o}$ - 150

<BAP = $30^{o}$

Apply the Sine rule to determine the value of PB, such that;

$\frac{AP}{Sin B}$ = $\frac{BP}{Sin30}$

$\frac{8}{Sin30}$ = $\frac{BP}{Sin 30}$

BP = $\frac{8*Sin30}{Sin 30}$

= $\frac{4}{0.5}$

BP = 8.0

Therefore, the value of BP = 8 units.

For more clarifications on applications of the Sine rule, visit: brainly.com/question/15018190

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1 year ago

The points (2, 3) and (6, 11) lie on a line. What is the slope of this line?

AveGali [126]

Answer:

8/4

Step-by-step explanation:

to do slope is y2 - y1 over x2 - x1, so it would be 11 - 3 for 8, then 6 - 2, leaving 8/4.

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2 years ago

If the complex number x = 3 + bi and |x|2 = 13, which is a possible value of b?

Naddika [18.5K]

Answer:

A: 2

Step-by-step explanation:

EDGE 2021

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2 years ago

Elizabeth attempts a field goal by kicking a football from the ground with an initial vertical

sammy [17]

$\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in feet} \\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{64}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{0\qquad \textit{from the ground}}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf h(t)=-16t^2+64t+0\implies h(t)=-16t^2+64t\implies \stackrel{\textit{hits the ground}~\hfill }{0=-16t^2+64t} \\\\\\ 0=-16t(t-4)\implies t= \begin{cases} 0\\ \boxed{4} \end{cases}$

Check the picture below, it hits the ground at 0 feet, where it came from, the ground, and when it came back down.

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2 years ago