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3 years ago

Which equation below represents the equation shown in the table

Mathematics

1 answer:

MrRa [10]3 years ago

7 0

Answer:

Step-by-step explanation: the answer is A because it is obvious that the other 4 arent possible equations

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Need reasons answer for this please!

Ugo [173]

Step-by-step explanation:

1.Given .

2.Being vertically opposite angle.

3.Because sum of all interior angle of a triangle is always 180°.

4.From the statement of 3 number.

5.Because angle 4 = angle 3 and angle 6= angle 1 .

6.(suppose ,< is angle sign) solving this equation,

m<1 + m<2+ m<3 = m<3 + m<5 + m<1

3 0

3 years ago

h(n)=-13nh(n)=−13nh, left parenthesis, n, right parenthesis, equals, minus, 13, n Complete the recursive formula of h(n)h(n)h, l

mr Goodwill [35]

Answer:

h(1)=-13

$h(n)=h(n-1)-13, n\geq 2$

Step-by-step explanation:

We are given that

$h(n)=-13n$

We have to find the recursive formula of h(n).

Substitute n=1

$h(1)=-13$

n=2

$h(2)=-13-13=-26=h(1)-13$

n=3

$h(3)=-13(3)=-39=-26-13=h(2)-13$

$h(4)=-13(4)=-52=-39-13=h(3)-13$

$h(n)=h(n-1)-13$

Therefore, the recursive formula is given by

h(1)=-13

$h(n)=h(n-1)-13, n\geq 2$

5 0

4 years ago

Ok please help me!!!!!!!

docker41 [41]

Answer:

keep it a bean bro I can't even see the question

Step-by-step explanation:

7 0

3 years ago

Forever Fitness charges $50 a month and $15 for each exercise class you take. Peachtree Gym charges $75 a month and $10 for each

Anarel [89]

Answer:

dont know sorry

Step-by-step explanation:

3 0

3 years ago

Find the exact value of cos theta, given that sin thetaequalsStartFraction 15 Over 17 EndFraction and theta is in quadrant II.

vova2212 [387]

Answer:

$cos \theta = -\frac{8}{17}$

Step-by-step explanation:

For this case we know that:

$sin \theta = \frac{15}{17}$

And we want to find the value for $cos \theta$ , so then we can use the following basic identity:

$cos^2 \theta + sin^2 \theta =1$

And if we solve for $cos \theta$ we got:

$cos^2 \theta = 1- sin^2 \theta$

$cos \theta =\pm \sqrt{1-sin^2 \theta}$

And if we replace the value given we got:

$cos \theta =\pm \sqrt{1- (\frac{15}{17})^2}=\sqrt{\frac{64}{289}}=\frac{\sqrt{64}}{\sqrt{289}}=\frac{8}{17}$

For our case we know that the angle is on the II quadrant, and on this quadrant we know that the sine is positive but the cosine is negative so then the correct answer for this case would be:

$cos \theta = -\frac{8}{17}$

5 0

3 years ago