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sergejj [24]
1 year ago
6

I wanted to know if my answer is correct and if it isn't please correct it.

Mathematics
2 answers:
dybincka [34]1 year ago
8 0

Answer:

Your answer is almost correct. The answer would be 10

Step-by-step explanation:

k0ka [10]1 year ago
4 0

Answer:10

Step-by-step explanation:

Use the distance formula to determine the distance between two points.

the answer will be 10

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Find the EQUATION of the line through (-10,-12) and (-15,3). Use the form y=mx+b.​
vovikov84 [41]
Y=Mx+b

Y=-3x-42

Please mark brainliest. Thank you
8 0
2 years ago
Write the complex number 1 +i/1 -i into polar form​
IRINA_888 [86]

Answer:

cos(pi/2) + i sin (pi/2)

Step-by-step explanation:

(1+i)/ (1-i)

Multiply top and bottom by (1+i)

(1+i) (1+i)/ (1-i)(1+i)

Foil the top (1+i)(1+i) = 1 +2i +i^2 = 1 +2i -1 = 2i

Foil the bottom ( 1+i)(1-i) = 1 - i^2 = 1 - (-1) =2

2i/2 = i

the magnitude is (0^2+1^2) = 1

The angle is tan^-1( 1/0) = pi/2

cos(pi/2) + i sin (pi/2)

7 0
3 years ago
If f(x)=2x-5, which expression represents f(x+1)? A) 2x-3, B) 2x-4, C)2x-5, D)2x+7
Dovator [93]
F(x)=x^2+2x+1 & g(x)=3(x+1)^2
now, f(x)+g(x)
=x^2+2x+1+3(x+1)^2
=x^2+2x+1+3(x^2+2x+1)
=x^2+2x+1+3x^2+6x+3
=4x^2+8x+4<===answer(c)
next:
f(x)=x^2-1 & g(x)=x+3
now, f(g(x))=(x+3)^ -1
=x^2+6x+9-1
=x^2+6x+8<====answer(b)
i solve two of ur problems.
now try the 3rd one that is similar to no. 1
and try the last two urself.
5 0
3 years ago
I need help ASAP due rn!
coldgirl [10]
There are no real solutions
5 0
3 years ago
NO LINKS!!! Part 2: Find the Lateral Area, Total Surface Area, and Volume. Round your answer to two decimal places.​
slava [35]

Answer:

<h3><u>Question 7</u></h3>

<u>Lateral Surface Area</u>

The bases of a triangular prism are the triangles.

Therefore, the Lateral Surface Area (L.A.) is the total surface area excluding the areas of the triangles (bases).

\implies \sf L.A.=2(10 \times 6)+(3 \times 6)=138\:\:m^2

<u>Total Surface Area</u>

Area of the isosceles triangle:

\implies \sf A=\dfrac{1}{2}\times base \times height=\dfrac{1}{2}\cdot3 \cdot \sqrt{10^2-1.5^2}=\dfrac{3\sqrt{391}}{4}\:m^2

Total surface area:

\implies \sf T.A.=2\:bases+L.A.=2\left(\dfrac{3\sqrt{391}}{4}\right)+138=167.66\:\:m^2\:(2\:d.p.)

<u>Volume</u>

\sf \implies Vol.=area\:of\:base \times height=\left(\dfrac{3\sqrt{391}}{4}\right) \times 6=88.98\:\:m^3\:(2\:d.p.)

<h3><u>Question 8</u></h3>

<u>Lateral Surface Area</u>

The bases of a hexagonal prism are the pentagons.

Therefore, the Lateral Surface Area (L.A.) is the total surface area excluding the areas of the pentagons (bases).

\implies \sf L.A.=5(5 \times 6)=150\:\:cm^2

<u>Total Surface Area</u>

Area of a pentagon:

\sf A=\dfrac{1}{4}\sqrt{5(5+2\sqrt{5})}a^2

where a is the side length.

Therefore:

\implies \sf A=\dfrac{1}{4}\sqrt{5(5+2\sqrt{5})}\cdot 5^2=43.01\:\:cm^2\:(2\:d.p.)

Total surface area:

\sf \implies T.A.=2\:bases+L.A.=2(43.01)+150=236.02\:\:cm^2\:(2\:d.p.)

<u>Volume</u>

\sf \implies Vol.=area\:of\:base \times height=43.011193... \times 6=258.07\:\:cm^3\:(2\:d.p.)

8 0
2 years ago
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