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4 years ago

What is y=x-4 and y=4x-10

Mathematics

2 answers:

PolarNik [594]4 years ago

6 0

Answer:

I got an answer of X = 2

Step-by-step explanation:

I used subtituion and got 2

mario62 [17]4 years ago

6 0

Answer:

Step-by-step explanation:

I've already answered this question.

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What is the gcf of 27 and 36

mestny [16]

The answer would be 9. 9x3=27 and 9x4=36. Although 3 fits into both, 9 is the BIGGER number that also fits into both of the numbers shown in the problem.

5 0

3 years ago

Read 2 more answers

MATH HELP PLZ!!!

RoseWind [281]

Answer:

a) $tan (157.5) = \frac{1-cos 315}{sin315}$

$sin (165) =\sqrt{ \frac{1-cos (330) }{2}}$

$sin^{2} (157.5) = \frac{1-cos (315) }{2}$

cos 330° = 1- 2 sin² (165°)

Step-by-step explanation:

Step(i):-

By using trigonometry formulas

cos2∝ = 2 cos² ∝-1

cos∝ = 2 cos² ∝/2 -1

1+ cos∝ = 2 cos² ∝/2

$cos^{2} (\frac{\alpha }{2}) = \frac{1+cos\alpha }{2}$

cos2∝ = 1- 2 sin² ∝

cos∝ = 1- 2 sin² ∝/2

$sin^{2} (\frac{\alpha }{2}) = \frac{1-cos\alpha }{2}$

Step(i):-

Given

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

we know that trigonometry formulas

$sin\alpha = 2sin(\frac{\alpha }{2} )cos(\frac{\alpha }{2} )$

1- cos∝ = 2 sin² ∝/2

Given

$tan(\frac{\alpha }{2} ) = \frac{sin(\frac{\alpha }{2} )}{cos(\frac{\alpha }{2}) }$

put ∝ = 315

$tan(\frac{315}{2} ) = \frac{sin(\frac{315 }{2} )}{cos(\frac{315 }{2}) }$

multiply with ' 2 sin (∝/2) both numerator and denominator

$tan (\frac{315}{2} )= \frac{2sin^{2}(\frac{315)}{2} }{2sin(\frac{315}{2} cos(\frac{315}{2}) }$

Apply formulas

$sin\alpha = 2sin(\frac{\alpha }{2} )cos(\frac{\alpha }{2} )$

1- cos∝ = 2 sin² ∝/2

now we get

$tan (157.5) = \frac{1-cos 315}{sin315}$

$sin^{2} (\frac{\alpha }{2}) = \frac{1-cos\alpha }{2}$

put ∝ = 330° above formula

$sin^{2} (\frac{330 }{2}) = \frac{1-cos (330) }{2}$

$sin^{2} (165) = \frac{1-cos (330) }{2}$

$sin (165) =\sqrt{ \frac{1-cos (330) }{2}}$

c )

$sin^{2} (\frac{\alpha }{2}) = \frac{1-cos\alpha }{2}$

put ∝ = 315° above formula

$sin^{2} (\frac{315 }{2}) = \frac{1-cos (315) }{2}$

$sin^{2} (157.5) = \frac{1-cos (315) }{2}$

cos∝ = 1- 2 sin² ∝/2

put ∝ = 330°

$cos 330 = 1 - 2sin^{2} (\frac{330}{2} )$

cos 330° = 1- 2 sin² (165°)

3 0

3 years ago

Number of factor of a + b whole square is

Morgarella [4.7K]

Answer:

(a+b)² = a² + b² + 2ab

7 0

4 years ago

Find all real values of t that satisfy the equation t^2 = 324. If you find more than one, then list your values in increasing or

vredina [299]

Answer:

t = -18, 18

Step-by-step explanation:

Take the square root of both sides of the equation.

... t = ±√324 = ±18

4 0

4 years ago

Compute the differential of surface area for the surface S described by the given parametrization.

AysviL [449]

With $S$ parameterized by

$\vec r(u,v)=\langle e^u\cos v,e^u\sin v,uv\rangle$

the surface element $\mathrm dS$ is

$\mathrm dS=\left\|\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

We have

$\dfrac{\partial\vec r}{\partial u}=\langle e^u\cos v,e^u\sin v,v\rangle$

$\dfrac{\partial\vec r}{\partial v}=\langle -e^u\sin v,e^u\cos v,u\rangle$

with cross product

$\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=\langle ue^u\sin v-ve^u\cos v,-ve^u\sin v-ue^u\cos v,e^{2u}\cos^2v+e^{2u}\sin^2v\rangle$

$\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=\langle e^u(u\sin v-v\cos v),-e^u(v\sin v+u\cos v),e^{2u}\rangle$

with magnitude

$\left\|\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right\|=\sqrt{e^{2u}(u\sin v-v\cos v)^2+e^{2u}(v\sin v+u\cos v)^2+e^{4u}}$

$\left\|\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right\|=e^u\sqrt{u^2+v^2+e^{2u}}$

So we have

$\mathrm dS=\boxed{e^u\sqrt{u^2+v^2+e^{2u}}\,\mathrm du\,\mathrm dv}$

8 0

3 years ago