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2 years ago

What is the value of x?And the perimeter and area?

Mathematics

1 answer:

Murrr4er [49]2 years ago

7 0

Answer:

Step-by-step explanation:

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How can you tell if an inequality will have a solid line

Virty [35]

Answer:

b. the symbol has or equal to

Step-by-step explanation:

Required

When does an inequality have a solid line

The question is pretty straightforward and the answer is (b) which means that when the symbol has or equal to.

This implies that, the symbol could be:

> or = i.e. $\ge$

and it could also be:

< or = i.e. $\le$

Take for instance:

$x + 1 \ge 2$ and $y- 5\le 10$

<em>The above expressions will have a solid line</em>

6 0

3 years ago

Please help it due rn

leonid [27]

Answer:

B) (1/2, -8)

Step-by-step explanation:

(1, -6) and (0, -10)

Midpoint formula:

((x1+x2)/2, (y1+y2)/2)

Solving for x:

(x1+x2)/2

(1 + 0)/2

1/2

Solving for y:

(y1+y2)/2

(-6-10)/2

(-16)/2

-8

8 0

2 years ago

Choose the best system of equations to represent the situation and determine the labels for the variables:

7nadin3 [17]

Answer B.
X represents the number of years grown
Y represents the total number of feet grown

4 0

3 years ago

Read 2 more answers

Triangle ABC is such that AB = 3cm , BC = 4cm , angle ABC = 120 and BAC = theta : Write down in terms of theta , and expression

lakkis [162]

1)inside the triangule is 180º .There are 3 angules, one with 120º, theta and
angle ACB, lets call it x.

x + theta + 120 = 180
x + theta= 180-120
x + theta= 60
x= 60- theta
angle ACB= 60- theta

8 0

3 years ago

Find the solution of the initial value problem<br><br> dy/dx=(-2x+y)^2-7 ,y(0)=0

Leokris [45]

Substitute $v(x)=-2x+y(x)$ , so that $\dfrac{\mathrm dv}{\mathrm dx}=-2+\dfrac{\mathrm dy}{\mathrm dx}$ . Then the ODE is equivalent to

$\dfrac{\mathrm dv}{\mathrm dx}+2=v^2-7$

which is separable as

$\dfrac{\mathrm dv}{v^2-9}=\mathrm dx$

Split the left side into partial fractions,

$\dfrac1{v^2-9}=\dfrac16\left(\dfrac1{v-3}-\dfrac1{v+3}\right)$

so that integrating both sides is trivial and we get

$\dfrac{\ln|v-3|-\ln|v+3|}6=x+C$

$\ln\left|\dfrac{v-3}{v+3}\right|=6x+C$

$\dfrac{v-3}{v+3}=Ce^{6x}$

$\dfrac{v+3-6}{v+3}=1-\dfrac6{v+3}=Ce^{6x}$

$\dfrac6{v+3}=1-Ce^{6x}$

$v=\dfrac6{1-Ce^{6x}}-3$

$-2x+y=\dfrac6{1-Ce^{6x}}-3$

$y=2x+\dfrac6{1-Ce^{6x}}-3$

Given the initial condition $y(0)=0$ , we find

$0=\dfrac6{1-C}-3\implies C=-1$

so that the ODE has the particular solution,

$\boxed{y=2x+\dfrac6{1+e^{6x}}-3}$

5 0

3 years ago