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3 years ago

PLZ HELP! I need to turn this in urgently <3

Mathematics

1 answer:

andriy [413]3 years ago

7 0

Answer:

Step-by-step explanation:

This is a 180° line and it would be 180-89=91.

This isn't much but hope it helps .^.

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PLZ HELP!!!<br> I will give you brainliest

Artist 52 [7]

Answer:

141

Step-by-step explanation:

CD = | 4-(-2)| = 6

DB = |3-(-7)| = 10

area of a right triangle

A = (leg 1 x leg2)/2 = (6 x 10)/2 = 30

EB = |3-(-3)| = 6

EF = |-8-(-2)| = 6

area of a square = side^2

A = 6^2 = 36

segment G(and a point with same x coordinate of H)

|8-3| = 5

segment H(and the same point)

|-10-(-8)| = 2

A = (5x2)/2 = 5

segment I(and the same point)

|-8-6| = 14

Area of a rectangle

A = base x width

A = (5 x 14) = 70

total area = 70 + 5 + 36 + 30 = 141

6 0

3 years ago

Gcf of -15xyz, -55xy²,- 55yz

GalinKa [24]

Answer:

-5y

Step-by-step explanation:

15 = 3×5

55 = 5×11

-15xyz = -1 × 3 × 5 × x × y × z

-55xy² = -1 × 5 × 11 × x × y²

-55yz = -1 × 5 × 11 × y × z

HCF = multiply all common factors with the lowest power amongst all 3 expressions

HCF = -1 × 5 × y = -5y

3 0

1 year ago

Write an equation of the perpendicular bisector of the segment with end points M(1,5) and N(7,-1)

vitfil [10]

The perpendicular bisector of the segment passes through the midpoint of this segment. Thus, we will initially find the midpoint P:

$P=\dfrac{(1,5)+(7,-1)}{2}=\dfrac{(8,4)}{2}=(4,2)$

Now, we will calculate the slope of the segment support line (r). After this, we will use the fact that the perpendicular bisector (p) is perpendicular to r:

$m_r=\dfrac{\Delta y}{\Delta x}=\dfrac{5-(-1)}{1-7}=\dfrac{6}{-6}\iff m_r=-1$

$p\perp r\Longrightarrow m_p\cdot m_r=-1\Longrightarrow m_p\cdot(-1)=-1\iff m_p=1$

We can calculate the equation of p by using its slope and its point P:

$y-y_P=m_p(x-x_P)\\\\
y-2=1\cdot(x-4)\\\\
y-2=x-4\\\\
\boxed{p:~~y=x-2}$

4 0

3 years ago

Simplify (square root 2)( square root of 2^3)

steposvetlana [31]

Answer:

$\large\boxed{(\sqrt2)(\sqrt{2^3})=4}$

Step-by-step explanation:

$(\sqrt2)(\sqrt{2^3})\qquad\text{use}\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\=\sqrt{2\cdot2^3}\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=\sqrt{2^{1+3}}=\sqrt{2^4}\\\\(1)=\sqrt{2^{2\cdot2}}\qquad\text{use}\ (a^n)^m=a^{nm}\\\\=\sqrt{(2^2)^2}\qquad\text{use}\ \sqrt{a^2}=a\ \text{for}\ a\geq0\\\\=2^2=4\\\\(2)=\sqrt{2\cdot2\cdot2\cdot2}=\sqrt{16}=6\ \text{because}\ 4^2=16$

8 0

3 years ago

I need help with this question it's really hardd

Vsevolod [243]

1st one on second row I think

7 0

3 years ago