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3 years ago

Solve the equation. Check the solution. -2/x+4 =4/x+3

Mathematics

2 answers:

mrs_skeptik [129]3 years ago

5 0

Answer:

option A is the answear

jasenka [17]3 years ago

4 0

Answer:B

Step-by-step explanation:

Looked it up

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What is log(4k-5)=log(2k-1)?

Paha777 [63]

Log(4k - 5) = log(2k - 1)
log(4k) - log(5) = log(2k) - log(1)
0.6020599913k - 0.6989700043 = 0.3010299957k - 0
0.6020599913k - 0.6989700043 = 0.3010299957k
-0.6020599913k -0.6020599913k
 -0.6989700043 = -0.3010299957k
 -0.3010299957 -0.3010299957
 2.321928094 = k

7 0

3 years ago

First choose which of the following statements are correct. If there is more than one reason why the system is not in echelon fo

tensa zangetsu [6.8K]

Answer:

D. The system is in echelon form.

Step-by-step explanation:

Check attachment

3 0

4 years ago

How do you do this question?

daser333 [38]

Answer:

B. 1/2

Step-by-step explanation:

$\lim_{z \to 0} \frac{g(z)e^{-z}-3}{z^{2}-2z}$

If we plug in 0 for z, we get 0/0. Apply l'Hopital's rule.

$\lim_{z \to 0} \frac{-g(z)e^{-z}+g'(z)e^{-z}}{2z-2}$

Now when we plug in 0 for z, we get:

$\frac{-g(0)e^{0}+g'(0)e^{0}}{2(0)-2}\\\frac{-g(0)+g'(0)}{-2}\\\frac{-3+2}{-2}\\\frac{1}{2}$

4 0

3 years ago

Could someone help me with number 19

azamat

Solving #19

Take y-values from the graph

a) (g·f)(-1) = g(f(-1)) = g(1) = 4
b) (g·f)(6) = g(f(6)) = g(2) = 2
c) (f·g)(6) = f(g(6)) = f(5) = 1
d) (f·g)(4) = f(g(4)) = f(2) = -2

8 0

3 years ago

Es 18 de junio de 1815, las tropas napoleónicas se encuentran justo adelante, tu eres el ingeniero en balística y el general Wel

Airida [17]

Answer:

89.1° or -1.4°

Step-by-step explanation:

1. Location:

You are on the Mont-Saint-Jean escarpment, near the Belgian town of Waterloo.

The French troops are about 50 m below you and 1.2 km distant.

2. Finding the firing angle

Data:

R = 1200 m

u = 600 m/s

h = -50 m (the height of the target)

a = 9.8 m/s²

We have two conditions.

Horizontal distance

(1) 1200 = 600t cosθ

Vertical distance

(2) -50 = 600t sinθ - 4.9t²

Divide each side of (1) by 600cosθ.

$(3) \, t =\dfrac{2}{\cos \theta}$

Substitute (3) into (2)

$-50 = 600t \sin \theta - 4.9t^{2} = 600 \left( \dfrac{2}{\cos \theta} \right ) \sin \theta - 4.9 \left( \dfrac{2}{\cos \theta} \right )^{2}\\\\(4) \, -50 = 1200 \tan \theta - \dfrac{19.6}{\cos^{2} \theta}$

Recall that

(5) sec²θ = 1/cos²θ = tan²θ + 1

Substitute (5) into (4)

$-50 = 1200 \tan \theta - 19.6 \left(\tan^{2} \theta}+ 1\right )$

Set up a quadratic equation

$\begin{array}{rcl}-50 & = & 1200 \tan \theta - 19.6\tan^{2} \theta -19.6 \\0 & = & 1200 \tan \theta - 19.6\tan^{2} \theta + 30.4\\0 & =&19.6\tan^{2} \theta - 1200 \tan \theta - 30.4\\0 & =&\tan^{2} \theta - 61.224 \tan \theta - 1.551\\\end{array}$

Solve for θ

Use the quadratic formula.

tanθ = 61.249 or -0.025

θ = arctan(61.249) = 89.1° or

θ = arctan(-0.025) = -1.4°

3 0

3 years ago