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4 years ago

The exterior angles of a triangle measure x + 3, 2x + 1, and 3x - 4. Find the measure of each exterior angle.

Mathematics

1 answer:

jonny [76]4 years ago

5 0

The answer to the question is c

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Here are the ingredients for making pineapple sorbet for 6 people.

kvasek [131]

Answer: 1200 g pinapple

Step-by-step explanation:

800/2=400*3=1200

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4 years ago

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How do you write a linear equation given the slope and a point that

Serga [27]

Answer:

point slope form

Step-by-step explanation:

hi! we can use point slope form:

y-y1=m(x-x1)

where m is the slope, x1 is the x-coordinate of the given point, and y1 is the y-coordinate of the given point.

example:

slope is 4

ordered pair is (2,3)

y-3=4(x-2)

y-3=4x-8

y=4x-5

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3 years ago

Find four consecutive integers whose sum is 234.

Agata [3.3K]

Answer:

57, 58, 59, 60

Hope this Helps :)

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3 years ago

Reason Abstractly Find two integers whose sum is -7 and whose

Travka [436]

Step-by-step explanation:

Let's represent the two integers with the variables $x$ and $y$ .

From the problem statement, we can create the following two equations:

$x + y = -7$

$xy = 12$

With the first equation, we can subtract $y$ from both sides to isolate the $x$ variable to the left-hand side:

$x = -7 - y$

Now that we have a value for $x$ , we can plug it into the second equation and solve for $y$ :

$(-7 - y)y = 12$

$-7y - y^{2} = 12$

Now, let's move everything to one side of the equation:

$y^{2} + 7y + 12 = 0$

Factoring this quadratic will give us two values for $y$ :

$(y + 4)(y + 3) = 0$

$y = -3, -4$

Since we now know $y = -3, -4$ , we can plug this back into either of the original equations to get a value for $x$ , which will be $x = -4, -3$ .

So the two numbers that sum to $-7$ and have a product of $12$ are $-3, -4$ .

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4 years ago

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I'll mark brainliest just tell me how to.

IRINA_888 [86]

Answer:

a = 12

b = 2

c = 11

Step-by-step explanation:

$\frac{( {x}^{5}y {z}^{4} )^{3} }{ {x}^{3} yz} = {x}^{a} {y}^{b} {z}^{c} \\ \\ \frac{ {x}^{5 \times 3}y^{3} {z}^{4 \times 3} }{ {x}^{3} yz} = {x}^{a} {y}^{b} {z}^{c} \\ \\ \frac{ {x}^{15}y^{3} {z}^{12} }{ {x}^{3} yz} = {x}^{a} {y}^{b} {z}^{c} \\ \\ {x}^{15 - 3} {y}^{3 - 1} {z}^{12 - 1} = {x}^{a} {y}^{b} {z}^{c} \\ \\ {x}^{12} {y}^{2} {z}^{11} = {x}^{a} {y}^{b} {z}^{c} \\ \\ equating \: like \: terms \: from \: both \: sides \\ \\ {x}^{12} = {x}^{a} \: \implies \: a = \boxed{ 12}\\ \\ {y}^{2} = {y}^{b} \: \implies \: b = \boxed{ 2} \\ \\ {z}^{11} = {z}^{c} \: \implies \: c = \boxed{ 11}$

3 0

3 years ago