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4 years ago

3

Mathematics

1 answer:

d1i1m1o1n [39]4 years ago

3 0

Answer: C

Step-by-step explanation: I just know the answer

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HURRY SIS (perpendicular,parallel,or none)

Aleksandr-060686 [28]

Answer:

perpendicular

Step-by-step explanation:

because the opp of 3 is $-\frac{1}{3}$

Red

4 0

4 years ago

Use the change of base formula to compute log5 9.

Mekhanik [1.2K]

Let log₅9 = x
Then
5ˣ = 9

Take log to base 10 of both sides.
log₁₀5ˣ = log₁₀9
x log₁₀5 = log₁₀9

Therefore
log₅9 = log₁₀9/log₁₀5
= 0.9542/0.699
= 1.3652

Answer: 1.365 (nearest thousandth)

5 0

3 years ago

A poster is to have an area of 210 in2 with 1 inch margins at the bottom and sides and a 2 inch margin at the top. find the exac

slavikrds [6]

Let

x-------> the length side of the poster

y-------> the width side of the poster

we know that

Area of the poster is equal to

$210=x*y\\\\ y=\frac{210}{x}$ -------> equation $1$

Area of the printed area is equal to

$Aprinted=(x-2)*(y-3)\\ Aprinted=xy-3x-2y+6$ ----> equation $2$

substitute equation $1$ in equation $2$

$Aprinted=[210]-3x-2*[\frac{210}{x}]+6$

using a graph tool

Find the vertex of the graph

The vertex of the graph is the point that represent the value of x for the largest printed area

see the attached figure

the vertex is the point $(11.83,145)$

the largest printed area is $145 in^{2}$

$x=11.83 in$

find the value of y

$y=\frac{210}{x} \\ \\ y=\frac{210}{11.83} \\ \\ y=17.75in$

therefore

the answer is

The dimensions of the poster are $11.83 in x 17.75 in$

8 0

3 years ago

Read 2 more answers

Express the area of an equilateral triangle as a function of the length of a side x.

VARVARA [1.3K]

Xroot3/2 is the answer

8 0

4 years ago

Please Help Quick!! Need full work shown!

kramer

Answer:

limit does not exists.

Step-by-step explanation:

From the given graph it is clear that the function approaches to 4 when x is approaches to 2 from left. So,

$\lim_{x\to 2^-}f(x)=4$

The function approaches to -4 when x is approaches to 2 from right. So,

$\lim_{x\to 2^+}f(x)=-4$

Now,

$4\neq -4$

$\lim_{x\to 2^-}f(x)\neq \lim_{x\to 2^+}f(x)$

Since, left hand limit is not equal to right hand limit, therefore, limit does not exists.

3 0

4 years ago