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3 years ago

1+2g=-7 how do you do this?

Mathematics

2 answers:

blsea [12.9K]3 years ago

8 0

Answer:

you would - 1 to -7 and get -8 then divide -8 by -2 than it would be 4

Step-by-step explanation:

s2008m [1.1K]3 years ago

7 0

I like my equations to have the variable before the constant.

2g+1=-7

Subtract 1 from the left side. Remember, whatever you do on one side, you MUST do it on the other.

2g=-8

Divide 2 from BOTH sides.

g=-4

Hope this helps!

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A rectangle has a length of 10 cm and a perimeter of 46 cm.

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B. since there are two sides that are 10, that equals 20. the other two sides are unknown, making the width 2w

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How can you determine whether two circles are congruent?

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If the diameter or radius of one circle is the same as another circle they are congruent.

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The glass fish tank does not have a cover. Find the total area of the glass panels used to make the tank.

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You will find the area of the 5 surfaces made of glass. There is a front, back, 2 sides, and the bottom. Each face is in the shape of a rectangle, so you will multiply the length by the width.
1. Front- 24 x 27 =648
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4 years ago

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Jan'ai was asked to determine the minimum for a function with zeros located at –1 and 5, which also has a y-intercept of (0, –25

sergey [27]

The first error in Jan'ai's work in determining the considered function is given by: Option D: She incorrectly determined the x-coordinate of the vertex.

<h3>What are the coordinates of vertex for a quadratic function?</h3>

For a quadratic function of the form $y = ax^2 + bx + c$ , its vertex form is obtained as:

$y = ax^2 + bx + c\\y =a(x^2 + bx/a) + c\\y = a(x^2 + 2(b/2a)x + (b/2a)^2 -(b/2a)^2 )+ c\\y = a(x^2 + 2(b/2a)x + (b/2a)^2) -a \times (b/2a)^2 + c\\\\y = a(x+b/2a)^2 - a \times (b/2a)^2 + c$

For the form $y = a(x-h)^2 + k$ , the vertex has coordinates (h, k)

Thus, for the obtained equation $y = a(x+b/2a)^2 - a \times (b/2a)^2 + c$ , we get the coordinates of vertex as:

$h = -b/2a$ , $k = c - a\times(b/2a)^2$

Thus, the coordinates of vertex of $y = ax^2 + bx + c$ is:

$(h,k) = (-b/2a, c - a \times (b/2a)^2 )$

The missing steps of work of Jan'ai are:

Begin to write a function in factored form. $f(x) = a(x+1)(x-5)$
Substitute x = 0, y = f(x) = -25 to determine a. $-25 = a(0+1)(0-5)$
Simplify and solve to find a. $a=5$
Rewrite the function. $f(x) = 5(x+1)(x-5)$
Rewrite in standard form. $f(x) = 5x^2-20x-25$
Find the x-coordinate of the vertex. $x = -20/2(5) = -20/10; x = -2$
Find the y-coordinate of the vertex.

$y = 5x^2-20x-25$

$y = 5(-2)^2-20(-2)-25$

$y = 35$

so (-2,35) is the coordinate of the vertex, which denotes the minimum.

So, as we see, in the 5th step, Jan'ai had the quadratic function $f(x) = 5x^2-20x-25$ ,

Comparing this to $f(x) = ax^2 + bx + c$ , we get a = 5, b = -20, c = -25

The vertex's x-coordinate will be on -b/2a = -(-20)/ 2(5) = 20/10 = 2

But Jan'ai didn't putted that negative sign before b. in the 6th step.

Thus, the first error in Jan'ai's work in determining the considered function is given by: Option D: She incorrectly determined the x-coordinate of the vertex.

Learn more about the vertex form of a quadratic equation here:

brainly.com/question/9912128

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3 years ago

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loris [4]

Answer:

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Step-by-step explanation:

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