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jeka94
3 years ago
13

Can someone help me please.

Mathematics
1 answer:
slega [8]3 years ago
5 0
6^log(216) is the answer
You might be interested in
-6(1 + 2x) + 6x = 6(-3x - 3)
Maksim231197 [3]

Answer:

x = - 1

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

PLEASE MARK BRAINLIEST!!

7 0
3 years ago
Fuel x costs $2 per gallon and fuel y costs $3 per gallon. You have at most $18 to spend on fuel. Write a system of linear inequ
Sloan [31]

The answer is :

x + y ≤ 14

2x + 3y ≤ 18

3 0
3 years ago
In Exercises 40-43, for what value(s) of k, if any, will the systems have (a) no solution, (b) a unique solution, and (c) infini
svet-max [94.6K]

Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

The system cannot have unique solution.

Step-by-step explanation:

We have the following system of equations

x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2

The augmented matrix is

\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]

The reduction of this matrix to row-echelon form is outlined below.

R_2\rightarrow R_2-R_1

\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]

R_3\rightarrow R_3-2R_1

\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]

R_3\rightarrow R_3-R_2

\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]

The last row determines, if there are solutions or not. To be consistent, we must have k such that

k^2-k-2=0

\left(k+1\right)\left(k-2\right)=0\\k=-1,\:k=2

Case k = −1:

\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]

If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]

This gives the infinite many solution.

5 0
3 years ago
Help pls :’( ASAPP!!!<br> “Complete the proof”
nata0808 [166]

1) \overline{AB} \cong \overline{CD}, \overline{AD} \cong \overline{CB}, \overline{AX} \perp \overline{BD}, \overline{CY} \perp\overline{BD} (given)

2) \overline{BD} \cong \overline{BD} (reflexive property)

3) \triangle ABD \cong \triangle ACDB (SSS)

4) \angle ADB \cong \angle CBY (CPCTC)

5) \angle CYB and \angle AXD are right angles (perpendicular lines form right angles)

6) \triangle CYB and \triangle AXD are right triangles (a triangle with a right angle is a right triangle)

7) \triangle AXD \cong \triangle CYB (HA)

8) \overline{AX} \cong \overline{CY} (CPCTC)

6 0
2 years ago
Find the area of the composite figure.
Roman55 [17]

Answer:

The Area of the composite figure would be 76.26 in^2

Step-by-step explanation:

<u>According to the Figure Given:</u>

Total Horizontal Distance = 14 in

Length = 6 in

<u>To Find :</u>

The Area of the composite figure

<u>Solution:</u>

Firstly we need to find the area of Rectangular part.

So We know that,

\boxed{ \rm \: Area  \:  of \:  Rectangle = Length×Breadth}

Here, Length is 6 in but the breadth is unknown.

To Find out the breadth, we’ll use this formula:

\boxed{\rm \: Breadth = total  \: distance - Radius}

According to the Figure, we can see one side of a rectangle and radius of the circle are common, hence,

\longrightarrow\rm \: Length \:  of \:  the  \: circle = Radius

  • Since Length = 6 in ;

\longrightarrow \rm \: 6 \: in   = radius

Hence Radius is 6 in.

So Substitute the value of Total distance and Radius:

  • Total Horizontal Distance= 14
  • Radius = 6

\longrightarrow\rm \: Breadth = 14-6

\longrightarrow\rm \: Breadth = 8 \: in

Hence, the Breadth is 8 in.

Then, Substitute the values of Length and Breadth in the formula of Rectangle :

  • Length = 6
  • Breadth = 8

\longrightarrow\rm \: Area \:  of  \: Rectangle = 6 \times 8

\longrightarrow \rm \: Area \:  of  \: Rectangle = 48 \: in {}^{2}

Then, We need to find the area of Quarter circle :

We know that,

\boxed{\rm Area_{(Quarter \; Circle) }  = \cfrac{\pi{r} {}^{2} }{4}}

Now Substitute their values:

  • r = radius = 6
  • π = 3.14

\longrightarrow\rm Area_{(Quarter \; Circle) } =  \cfrac{3.14 \times 6 {}^{2} }{4}

Solve it.

\longrightarrow\rm Area_{(Quarter \; Circle) } =  \cfrac{3.14 \times 36}{4}

\longrightarrow\rm Area_{(Quarter \; Circle) } =  \cfrac{3.14 \times \cancel{{36} } \: ^{9} }{ \cancel4}

\longrightarrow\rm Area_{(Quarter \; Circle)} =3.14 \times 9

\longrightarrow\rm Area_{(Quarter \; Circle) } = 28.26 \:  {in}^{2}

Now we can Find out the total Area of composite figure:

We know that,

\boxed{ \rm \: Area_{(Composite Figure)} =Area_{(rectangle)}+ Area_{ (Quarter Circle)}}

So Substitute their values:

  • \rm Area_{(rectangle)} = 48
  • \rm Area_{(Quarter Circle)} = 28.26

\longrightarrow \rm \: Area_{(Composite Figure)} =48 + 28 .26

Solve it.

\longrightarrow \rm \: Area_{(Composite Figure)} =\boxed{\tt 76.26 \:\rm in {}^{2}}

Hence, the area of the composite figure would be 76.26 in² or 76.26 sq. in.

\rule{225pt}{2pt}

I hope this helps!

3 0
2 years ago
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