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Mnenie [13.5K]
3 years ago
6

Please help asap!!!!!!!!!!!!!!!!!!! will give brainliest

Mathematics
1 answer:
PilotLPTM [1.2K]3 years ago
4 0

Answer:

no lo se

Step-by-step explanation:

44

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Find the surface area of the pyramid below. PLS HELPPP
Ksenya-84 [330]

Answer:

149.45 cm²

Step-by-step explanation:

Surface area of pyramid = ½Pl + BA

Perimeter of the base (P) = 7 + 7 + 7 = 21 cm

Slant height (l) = 12.2 cm

Area of base (BA) = ½*bh = ½*7*6.1 = 21.35 cm²

Plug in the values

Surface area = ½*21*12.2 + 21.35 = 149.45 cm²

5 0
3 years ago
What’s the chord BC?
Mashutka [201]

Answer:

  D)  3.8 cm

Step-by-step explanation:

There are several ways this problem can be solved. Maybe the easiest is to use the Law of Cosines to find angle BAC. Then trig functions can be used to find the length of the chord.

__

In triangle BAC, the Law of Cosines tells us ...

  a² = b² +c² -2bc·cos(A)

  A = arccos((b² +c² -a²)/(2bc)) = arccos((8² +6² -3²)/(2·8·6)) = arccos(91/96)

  A ≈ 18.573°

The measure of half the chord is AB times the sine of this angle:

  BD = 2(AB·sin(A)) ≈ 3.82222

The length of the common chord is about 3.8 cm.

_____

<em>Additional comment</em>

Another solution can be found using Heron's formula to find the area of triangle ABC. From that, its altitude can be found.

  Area ABC = √(s(s-a)(s-b)(s-c)) . . . . where s=(a+b+c)/2

  s=(3+8+6)/2 = 8.5

  A = √(8.5(8.5 -3)(8.5 -8)(8.5 -6)) = √54.4375 ≈ 7.64444

The altitude of triangle ABC to segment AC is given by ...

  A = 1/2bh

  h = 2A/b = 2(7.64444)/8 = 1.911111

BD = 2h = 3.822222

5 0
2 years ago
What is the surface area, in cm2, of the figure shown above?
Delvig [45]

I think it might be 14 not sure

4 0
3 years ago
Find the nonpermissible replacement for x in<br> this expression.<br> X-4/X-8
vitfil [10]

Answer:

8.

Step-by-step explanation:

x cannot be 8 because that would make the denominator zero.

8 0
3 years ago
What is the equation of the line that passes through the points (-2,2) and (0,5)
gregori [183]
\bf \begin{array}{lllll}&#10;&x_1&y_1&x_2&y_2\\&#10;%   (a,b)&#10;&({{ -2}}\quad ,&{{ 2}})\quad &#10;%   (c,d)&#10;&({{ 0}}\quad ,&{{ 5}})&#10;\end{array}&#10;\\\\\\&#10;% slope  = m&#10;slope = {{ m}}= \cfrac{rise}{run} \implies &#10;\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{5-2}{0-(-2)}\implies \cfrac{5-2}{0+2}&#10;\\\\\\&#10;\cfrac{3}{2}

\bf y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-2=\cfrac{3}{2}(x-(-2))\\&#10;\left. \qquad   \right. \uparrow\\&#10;\textit{point-slope form}&#10;\\\\\\&#10;y-2=\cfrac{3}{2}(x+2)\implies y-2=\cfrac{3}{2}x+3\implies y=\cfrac{3}{2}x+3+2&#10;\\\\\\&#10;y=\cfrac{3}{2}x+5
4 0
3 years ago
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