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3 years ago

Two values of x that are roots of xsquared+2x-6=0

Mathematics

2 answers:

xz_007 [3.2K]3 years ago

7 0

The two values of x are

1.65 , -3.65

Flura [38]3 years ago

4 0

Answer:

Two values of x that are :

$x=-\sqrt{7}-1$

$x=\sqrt{7}-1$

Step-by-step explanation:

Given= $x^2+2x-6=0$

To find = Two values of x

Solution :

$x^2+2x-6=0$

Adding and subtracting 1 in the equation :

$x^2+2x+1-1-6=0$

$(x^2+2\times x\times 1+1^2)-1-6=0$

$(x+1)^2-7=0$

$(x+1)^2=7$

$(x+1)=\pm \sqrt{7}$

Two values of x that are :

$x=-\sqrt{7}-1$

$x=\sqrt{7}-1$

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PLEASE HELP ASAP ILL GIVE BRAINLIEST!!!!!<br><br>find measure of arc MK

Gwar [14]

Answer:

Arc length MK = 15.45 units (nearest hundredth)

Arc measure = 58.24°

Step-by-step explanation:

Calculate the measure of the angle KLN (as this equals m∠KLM which is the measure of arc MK)

ΔKNL is a right triangle, so we can use the cos trig ratio to find ∠KLM:

$\sf \cos(\theta)=\dfrac{A}{H}$

where:

$\theta$ is the angle
A is the side adjacent the angle
H is the hypotenuse (the side opposite the right angle)

Given:

$\theta$ = ∠KLM
A = LN = 8
H = KL = 15.2

$\implies \sf \cos(KLM)=\dfrac{8}{15.2}$

$\implies \sf \angle KLM=\cos^{-1}\left(\dfrac{8}{15.2}\right)$

$\implies \sf \angle KLM=58.24313614^{\circ}$

Therefore, the measure of arc MK = 58.24° (nearest hundredth)

$\textsf{Arc length}=2 \pi r\left(\dfrac{\theta}{360^{\circ}}\right) \quad \textsf{(where r is the radius and}\:\theta\:{\textsf{is the angle)}$

Given:

r = 15.2
∠KLM = 58.24313614°

$\implies \textsf{Arc length MK}=2 \pi (15.2)\left(\dfrac{\sf \angle KLM}{360^{\circ}}\right)$

$\implies \textsf{Arc length MK}=\sf 15.45132428\:units$

6 0

2 years ago

Triangle ABC

Virty [35]

Answer:

15 cm

Step-by-step explanation:

Similar triangles are triangles which have the same shape but not the same size. This means their angle measures are equal but their side lengths are not instead they are proportional. To solve, we will set up a proportion and solve.

A proportion is two equal ratios set equal to each other. We form the ratios by finding corresponding sides (sides which match to each other on both triangles by their position). Out ratios will be one side of the small triangle over the corresponding side on the big triangle as shown below:

$\frac{BC}{EF} =\frac{AC}{DF}$ .