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3 years ago

15p+3(p-1)>3(2) exponent of 3 on two

Mathematics

1 answer:

Bas_tet [7]3 years ago

6 0

15p + 3(p-1) > 3(2^3)
15p + 3p - 3 > 3(8)
18p - 3 > 24
18p > 24 + 3
18p > 27
p > 27/18
p > 3/2 or 1 1/2

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The picture is incorrect but what is the correct value of x

Monica [59]

Start with -3x + 2 = -7
Subtract 2 from each side
-3x = -9
Divide each side by -3
x = 3

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3 years ago

Can someone help me please

svp [43]

Answer:

Step-by-step explanation:

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2 years ago

What effect would doubling all the dimensions of a triangular pyramid have on the volume of the pyramid?

lana66690 [7]

Answer:

The effect of doubling all the dimensions of a triangular pyramid will have on the volume of the pyramid is to increase it by 8 times.

Step-by-step explanation:

i) volume of triangular pyramid = $\frac{1}{3}$ $\times$ Area $\times$ height

= $\frac{1}{3}$ $\times$ (Base of triangle $\times$ perpendicular height of triangle) $\times$ height

ii) if we double all the dimensions then the three variables((Base of triangle, perpendicular height of triangle, height of pyramid) will be doubled

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2 years ago

Segments

Ipatiy [6.2K]

There are several ways two triangles can be congruent.

$\mathbf{AC = BD}$ congruent by SAS
$\mathbf{\angle ABC \cong \angle BAD}$ congruent by corresponding theorem

In $\mathbf{\triangle AOL}$ and $\mathbf{\triangle BOK}$ (see attachment), we have the following observations

$1.\ \mathbf{AO = DO}$ --- Because O is the midpoint of line segment AD

$2.\ \mathbf{BO = CO}$ --- Because O is the midpoint of line segment BC

$3.\ \mathbf{\angle AOB =\angle COD}$ ---- Because vertical angles are congruent

$4.\ \mathbf{\angle AOC =\angle BOD}$ ---- Because vertical angles are congruent

Using the SAS (side-angle-side) postulate, we have:

$\mathbf{AC = BD}$

Using corresponding theorem,

$\mathbf{\angle ABC \cong \angle BAD}$ ---- i.e. both triangles are congruent

The above congruence equation is true because:

2 sides of both triangles are congruent
1 angle each of both triangles is equal
Corresponding angles are equal

See attachment

Read more about congruence triangles at:

brainly.com/question/20517835

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2 years ago

Tan^2 A/1+cot^2 A + cot^2 A/1+tan^2 A=sec^2 A cosec^2 A-3

omeli [17]

$\frac{tan^2x}{1+cot^2x}+\frac{cot^2x}{1+tan^2x}=sec^2x\ cosec^2x-3\\\\L=\frac{tan^2x(1+tan^2x)+cot^2x(1+cot^2x)}{(1+cot^2x)(1+tan^2x)}=\frac{tan^2x+tan^4x+cot^2x+cot^4x}{1+tan^2x+cot^2x+tanxcotx}\\\\=\frac{tan^2x+cot^2x+tan^4x+cot^4x}{1+tan^2x+cot^2x+1}=\frac{tan^2x+2+cot^2x+tan^4x-2+cot^4x}{tan^2x+cot^2x+2}$

$=\frac{(tanx+cotx)^2+(tan^2x-cot^2x)^2}{(tanx+cotx)^2}=\frac{(tanx+cotx)^2}{(tanx+cotx)^2}+\frac{(tan^2x-cot^2x)^2}{(tanx+cotx)^2}\\\\=1+\frac{(tanx-cotx)^2(tanx+cotx)^2}{(tanx+cotx)^2}=1+(tanx-cotx)^2\\\\=1+tan^2x-2tanx\ cotx+cot^2x=tan^2x+cot^2x+1-2\\\\=\left(\frac{sinx}{cosx}\right)^2+\left(\frac{cosx}{sinx}\right)^2-1=\frac{sin^2x}{cos^2x}+\frac{cos^2x}{sin^2x}-1=\frac{sin^4x+cos^4x}{sin^2x\ cos^2x}-1$

$=\frac{(sin^2x)^2+2sin^2x\ cos^2x+(cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{(sin^2x+cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{1}{sin^2x\ cos^2x}-\frac{2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1}{sin^2x}\cdot\frac{1}{cos^2x}-2-1\\\\=cosec^2x\cdot sec^2x-3=sec^2x\ cosec^2x-3=R$

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3 years ago