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Gemiola [76]
2 years ago
10

4 (3×-1)=3+8×-11 2 (t+2)+5t=6t+11

Mathematics
1 answer:
babymother [125]2 years ago
3 0
X= -1 t= 7

4(3x-1)=3+8x-11
12x-4=3+8x-11
12x-8x=4+3-11
4x= -4
\4. \4
x= -1

2t+4+5t=6t+11
2t-6t+5t=-4+11
1t=7
\1. \1
t=7
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3 years ago
How do I find the vertex and the standard form? Plz explain
xenn [34]

Answer:

Vertex form

y = -4(x + 3)^2 + 10

Standard form;

y = -4x^2-24x - 26

Step-by-step explanation:

Mathematically, we have the vertex form as

y = a(x-h)^2 + k

(h,k) represents the vertex

We have h as -3 and k as 10

y = a(x+3)^2 + 10

To get a, we substitute any of the points

Let us use (-1,-6)

-6 = a(-1+3)^2 + 10

-6-10 = 4a

4a = -16

a = -16/4

a = -4

So we have the equation as;

y = -4(x+3)^2 + 10

For the standard form;

We expand the vertex form;

y = -4(x + 3)(x + 3) + 10

y = -4(x^2 + 6x + 9) + 10

y = -4x^2 - 24x -36 + 10

y = -4x^2 -24x -26

7 0
2 years ago
The sum of an acute angle and an obtuse angle is an acute angle always sometimes or never true ???​
xxTIMURxx [149]

Answer:

Never

Step-by-step explanation:

Adding together an obtuse angle (greater than 90°) and acute angle (less than 90°) should result in an obtuse angle every time

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xxTIMURxx [149]

Answer:

(1-\sqrt{2})a^2

Step-by-step explanation:

Consider irght triangle PRS. By the Pythagorean theorem,

PS^2=PR^2+RS^2\\ \\PS^2=a^2+a^2\\ \\PS^2=2a^2\\ \\PS=\sqrt{2}a

Thus,

MS=PS-PM=\sqrt{2}a-a=(\sqrt{2}-1)a

Consider isosceles triangle MSC. In this triangle

MS=MC=(\sqrt{2}-1)a.

The area of this triangle is

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Consider right triangle PTS. The area of this triangle is

A_{PTS}=\dfrac{1}{2}PT\cdot TS=\dfrac{1}{2}a\cdot a=\dfrac{a^2}{2}

The area of the quadrilateral PMCT is the difference in area of triangles PTS and MSC:

A_{PMCT}=\dfrac{(3-2\sqrt{2})a^2}{2}-\dfrac{a^2}{2}=\dfrac{(2-2\sqrt{2})a^2}{2}=(1-\sqrt{2})a^2

5 0
3 years ago
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