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3 years ago

Solve 24x+8x-11x= -7-14

Mathematics

2 answers:

kolbaska11 [484]3 years ago

8 0

24x+8x= 32x
32x-11x=21x
21x=-7-14
-7-14=-21
21x=-21
-1

astra-53 [7]3 years ago

5 0

Answer:

Step-by-step explanation:

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What is the cost price of 600 pens

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Answer:

600 *10=6000 if each of these are of 10

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3 years ago

Solve the system of equations by graphing.

valentinak56 [21]

The answer is A.

The way you know this is by graphing the first function.

The second function is in a different format, so you just have to isolate Y.

-1/3x + y = -1
y= 1/3x -1 OR y= -1 + 1/3x

both of the functions have the same slope, but different y-intercepts. therefore, they'll never touch. making them infinite solutions.

it's also A because both of the slopes are positive, the slopes in graph D are negative.

So, the answer is A.

Hope this helps! xx

4 0

3 years ago

Overline MD cong overline LS additional information is necessary to show that triangle MTD cong triangle LGS by SSS?

aivan3 [116]

Answer:

$TD \cong GS$

Step-by-step explanation:

See comment for complete question

Given:

$TM \cong GL$

$MD \cong LS$

Required

The information that shows $\triangle MTD \cong \triangle LGS$ by SSS

By SSS implies that, the three sides of both triangles are congruent

Already, we have:

$TM \cong GL$

$MD \cong LS$

The third side of $\triangle MTD$ is $TD$

The third side of $\triangle LGS$ is $GS$

So, for both to be congruent by SSS, the third sides must be congruent

i.e.

$TD \cong GS$

4 0

2 years ago

Graph the hyperbola using the transverse axis, vertices, and co-vertices:

Reptile [31]

See attachment for the graph of the hyperbola 12x^2 - 3y^2 - 108 = 0

<h3>How to graph the hyperbola?</h3>

The equation of the hyperbola is given as:

12x^2 - 3y^2 - 108 = 0

Start by calculating the transverse axis

So, we have:

<u>Transverse axis</u>

The vertices of the given hyperbola are (0, 0)

This means that

(h, k) = 0

Where

a = 3 and b = 6

The transverse axis is calculated as:

y = ±b/a(x - h) + k

So, we have:

y = ±6/3(x - 0) + 0

Evaluate the difference and sum

y = ±6/3x

Evaluate the quotient

y = ±2x

This means that the transverse axes are y = 2x and y =-2x

<u>The vertices</u>

In the above section, we have:

The vertices of the given hyperbola are (0, 0)

This means that

(h, k) = 0

<u>The co-vertices</u>

In the above section, we have:

The vertices of the given hyperbola are (0, 0)

This means that

(h, k) = 0

And

a = 3 and b = 6

The co-vertices are

(h - a, k) and (h + a, k)

So, we have:

(0 - 3, 0) and (0 + 3, 0)

Evaluate

(-3,0) and (3, 0)

See attachment for the graph of the hyperbola