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3 years ago

If a is a solution of abs(x+3)>4, then is a also a solution of x+3>4?

Mathematics

1 answer:

Alexandra [31]3 years ago

7 0

Answer:

Yes

Step-by-step explanation:

After solving for the two values of a, the second inequality gives a value inside the range of the first inequality.

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Select the correct answer.

lidiya [134]

Answer: y=2x+7

(this is a rearranged version of y-2x=7 from question)

Step-by-step explanation:

4x=5-2y

y-2x=7

i would rearrange the 2nd value into y intercept form y=mx+b

so move the -2x to the other side of the = sign by doing the opposite, +2x on both sides to isolate the y

y=2x+7

you didn't give any options for "correct answer" so i can't give

you A B C or D answers

8 0

1 year ago

Nshdjdfhdjdjddjjddjjddjdjdk

Allushta [10]

Answer:

yes

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Find the exact values of a) sec of theta b)tan of theta if cos of theta= -4/5 and sin<0

Gre4nikov [31]

Answer:

Using trigonometric ratio:

$\sec \theta = \frac{1}{\cos \theta}$

$\tan \theta = \frac{\sin \theta}{\cos \theta}$

From the given statement:

$\cos \theta = -\frac{4}{5}$ and sin < 0

⇒ $\theta$ lies in the 3rd quadrant.

then;

$\sec \theta = \frac{1}{-\frac{4}{5}} = -\frac{5}{4}$

Using trigonometry identities:

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

Substitute the given values we have;

$\sin \theta = \pm\sqrt{1-(\frac{-4}{5})^2 } =\pm\sqrt{1-\frac{16}{25}} =\pm\sqrt{\frac{25-16}{25}} =\pm \sqrt{\frac{9}{25} } = \pm\frac{3}{5}$

Since, sin < 0

⇒ $\sin \theta = -\frac{3}{5}$

now, find $\tan \theta$ :

$\tan \theta = \frac{\sin \theta}{\cos \theta}$

Substitute the given values we have;

$\tan \theta = \frac{-\frac{3}{5} }{-\frac{4}{5} } = \frac{3}{5}\times \frac{5}{4} = \frac{3}{4}$

Therefore, the exact value of:

(a)

$\sec \theta =-\frac{5}{4}$

(b)

$\tan \theta= \frac{3}{4}$

7 0

3 years ago

Verify identity list steps. Cot(t)(1-cos^2(t))=cos(t)sin(t)

storchak [24]

Remember: We have to work from either the LHS or the RHS.
(Left hand side or the Right hand side)

You should already know this:

$\huge{Cot(t) = \frac{1}{tan(t)} = \frac{1}{\frac{sin(t)}{cos(t)}} = 1\div \frac{sin(t)}{cos(t)} = 1\times \frac{cos(t)}{sin(t)}=\boxed{\frac{cos(t)}{sin(t)}}$

You should also know this:

$sin^2(t) + cos^2(t) = 1\\\\\boxed{sin^2(t)} = 1 - cos^2(t)$

So plugging in both of those into our identity, we get:

$\frac{cos(t)}{sin(t)}\cdot sin^2(t) = cos(t)\cdot sin(t)$

Simplify the denominator on the LHS (Left Hand Side)

We get:

$cos(t) \cdot sin(t) = cos(t) \cdot sin(t)$

LHS = RHS

Therefore, identity is verified.

4 0

3 years ago

The height of a cylinder is 5 centimeters. The diameter of the base of the cylinder is 16 centimeters. Find the lateral surface

soldier1979 [14.2K]

Answer:

Lateral surface area of the cylinder = $251.2 cm^2$