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3 years ago

Solve the inequality -x^2-4x+5>0

1 answer:

5 0

−x2−4x+5>0
Let's find the critical points of the inequality.
−x2−4x+5=0
(−x+1)(x+5)=0(Factor left side of equation)
−x+1=0 or x+5=0(Set factors equal to 0)
x=1 or x=−5
Check intervals in between critical points. (Test values in the intervals to see if they work.)
x<−5(Doesn't work in original inequality)
−5<x<1(Works in original inequality)
x>1(Doesn't work in original inequality)
Answer: −5<x<1

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Please answer quick!

marysya [2.9K]

Answer:

The correct option is B

(-1)(1/2)(-1)(1)

Step-by-step explanation:

First thing to notice is that there is are two brackets with negative values, which means that the result must also be positive (or have two brackets with negative values).

Looking at the options, we can screen out options A and D, they have three brackets with negative values, and can't be chosen.

It is now between options B and C.

Option B, by inspection is simply 1/2

Option C is 12/4 = 3

The problem itself is 12/35

12/24 = 1/2

12/36 = 1/3

Since 12/35 is about 1/3, it is closer to 1/2 than 3, so 1/2 is the best option.

6 0

3 years ago

The function h(t)=-4.92t^2+17.69t+575 is used to model an object being tossed from a tall building, where h(t) is the height in

nikklg [1K]

We have the following equation:
 h(t)=-4.92t^2+17.69t+575

 For the domain we have:
 We match zero:
-4.92t ^ 2 + 17.69t + 575 = 0
We look for the roots:
t1 = -9.16
t2 = 12.76
We are left with the positive root, so the domain is:
[0, 12.76]

For the range we have:
We derive the function:
h '(t) = - 9.84t + 17.69
We equal zero and clear t:
-9.84t + 17.69 = 0
t = 17.69 / 9.84
t = 1.80
We evaluate the time in which it reaches the maximum height in the function:
h (1.80) = - 4.92 * (1.80) ^ 2 + 17.69 * (1.80) +575
h (1.80) = 590.90
Therefore, the range is given by:
[0, 590.9]

Answer:
the domain and range are:
domain: [0, 12.76] range: [0, 590.9]

5 0

3 years ago

Read 2 more answers

Hey I need a step by step of this!!! Or can you just explain to me how to figure it out step by step!! either is fine

Umnica [9.8K]

Answer:

I believe the answer may be x= 24.5 I'm not certain though

Step-by-step explanation:

The angles inside the triangle must all add up to 180 so we can subtract 80 from 180 to get 98. Now I'm assuming the two undefined angles are equal because of the two angle bisectors that create the second triangle. If I'm correct in assuming that them we can divided 98 by 2 to get 49 which would be the measurement of both bottom angles. Since there's a line bisecting both of them we would then cut that number in half to get 24.5. I hope this helped, I'm not fully certain of this answer though.

4 0

3 years ago

Use the substitution method to solve the system of equations

PilotLPTM [1.2K]

You solve the substitution method to solve a system of equality by expressing one variable in terms of the other using one equation, and then plugging this expression in the other(s).

In this case, the first equation gives us a way to express n in terms of m. So, we can replace every occurrence of n in the second equation with the given formula.

The result is

$14m+2n=-8 \iff 14m+2(-7m-4)=-8 \iff 14m-14m-8=-8 \iff -8=-8$

So, the second equation turned to be an equality, i.e. an equation where both sides are the same.

This implies that the system has infinitely many solutions, because every couple $(n,m)$ such that $n=-7m-4$ is a solution to the system, because it satisfies both equations: the first is trivially satisfied, whereas the second is an identity, and as such is satisfied by any value of the variable.

3 0

3 years ago

Prove that: (b²-c²/a)CosA+(c²-a²/b)CosB+(a²-b²/c)CosC = 0

IRISSAK [1]

Prove that:

$\:\:\sf\:\:\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C=0$

Proof: 

We know that, by Law of Cosines,

$\sf \cos A=\dfrac{b^2+c^2-a^2}{2bc}$
$\sf \cos B=\dfrac{c^2+a^2-b^2}{2ca}$
$\sf \cos C=\dfrac{a^2+b^2-c^2}{2ab}$

Taking LHS

$\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C$

Substituting the value of cos A, cos B and cos C,

$\longmapsto\left(\dfrac{b^2-c^2}{a}\right)\left(\dfrac{b^2+c^2-a^2}{2bc}\right)+\left(\dfrac{c^2-a^2}{b}\right)\left(\dfrac{c^2+a^2-b^2}{2ca}\right)+\left(\dfrac{a^2-b^2}{c}\right)\left(\dfrac{a^2+b^2-c^2}{2ab}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2-a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2-b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2-c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2)-(b^2-c^2)(a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2)-(c^2-a^2)(b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2)-(a^2-b^2)(c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^4-c^4)-(a^2b^2-a^2c^2)}{2abc}\right)+\left(\dfrac{(c^4-a^4)-(b^2c^2-a^2b^2)}{2abc}\right)+\left(\dfrac{(a^4-b^4)-(a^2c^2-b^2c^2)}{2abc}\right)$

$\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2}{2abc}+\dfrac{c^4-a^4-b^2c^2+a^2b^2}{2abc}+\dfrac{a^4-b^4-a^2c^2+b^2c^2}{2abc}$

On combining the fractions,

$\longmapsto\dfrac{(b^4-c^4-a^2b^2+a^2c^2)+(c^4-a^4-b^2c^2+a^2b^2)+(a^4-b^4-a^2c^2+b^2c^2)}{2abc}$

$\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2+c^4-a^4-b^2c^2+a^2b^2+a^4-b^4-a^2c^2+b^2c^2}{2abc}$

Regrouping the terms,

$\longmapsto\dfrac{(a^4-a^4)+(b^4-b^4)+(c^4-c^4)+(a^2b^2-a^2b^2)+(b^2c^2-b^2c^2)+(a^2c^2-a^2c^2)}{2abc}$

$\longmapsto\dfrac{(0)+(0)+(0)+(0)+(0)+(0)}{2abc}$

$\longmapsto\dfrac{0}{2abc}$

$\longmapsto\bf 0=RHS$

LHS = RHS proved.

7 0

2 years ago