Which of these are the x- and y-intercepts of the graph?

Mathematics

1 answer:

Blizzard [7]3 years ago

7 0

Your answer will be B. Your x value is on -6 and y would be 0 at that point, making it (-6,0). Your y intercept is on the 3 and x is 0, making it (0,3).

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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

(2x3z2)3 <br> ------------------ <br> x3y4z2 * (x-4z3)

Komok [63]

The rules of exponents tell you to add exponents in the numerator and subtract those in the denominator. A power of a power causes the exponent to be multiplied.

$\dfrac{(2x^{3}z^{2})^{3}}{x^{3}y^{4}z^2x^{-4}z^{3}}=2^{3}x^{(3\cdot3-3-(-4))}y^{-4}z^{(2\cdot3-2-3)}\\\\=\dfrac{8x^{10}z}{y^{4}}$