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3 years ago

Which is a correct first step for solving this equation? 4x=7+3(2x−5)

Mathematics

2 answers:

Sholpan [36]3 years ago

6 0

The first step you would need to complete would be multiplying everything within the parentheses by 3.<span />

jeka943 years ago

3 0

Doing (2x-5) first is what

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Warm-Up

Serjik [45]

Answer:

using the formula

we replace the coordinates

the we calculate

the answer will be square 100 which will give us 10

5 0

3 years ago

Please answer all please

Firdavs [7]

Answer:

Step-by-step explanation:

The first parabola has vertex (-1, 0) and y-intercept (0, 1).

We plug these values into the given vertex form equation of a parabola:

y - k = a(x - h)^2 becomes

y - 0 = a(x + 1)^2

Next, we subst. the coordinates of the y-intercept (0, 1) into the above, obtaining:

1 = a(0 + 1)^2, and from this we know that a = 1. Thus, the equation of the first parabola is

y = (x + 1)^2

Second parabola: We follow essentially the same approach. Identify the vertex and the two horizontal intercepts. They are:

vertex: (1, 4)

x-intercepts: (-1, 0) and (3, 0)

Subbing these values into y - k = a(x - h)^2, we obtain:

0 - 4 = a(3 - 1)^2, or

-4 = a(2)². This yields a = -1.

Then the desired equation of the parabola is

y - 4 = -(x - 1)^2

7 0

3 years ago

Cameron had 30 minutes to do a three problem quiz. He spent 8 3/4 minutes on question A and 5 1/2 minutes on question B. How muc

Delvig [45]

He has 15 and 3/4 minutes left over

3 0

2 years ago

In the given figure △ABC ≅△DEC. Which of the following relations can be proven using CPCTC ?

Serhud [2]

Option B:

$\overline{A B}=\overline{D E}$

Solution:

In the given figure $\triangle A B C \cong \triangle D E C$ .

If two triangles are similar, then their corresponding sides and angles are equal.

By CPCTC, in $\triangle A B C \ \text{and}\ \triangle D E C$ ,

$\overline{AB }=\overline{DE}$ – – – – (1)

$\overline{B C}=\overline{EC}$ – – – – (2)

$\overline{ CA}=\overline{CD}$ – – – – (3)

$\angle ACB=\angle DCE$ – – – – (4)

$\angle ABC=\angle DEC$ – – – – (5)

$\angle BAC=\angle EDC$ – – – – (6)

Option A: $\overline{B C}=\overline{D C}$

By CPCTC proved in equation (2) $\overline{B C}=\overline{EC}$ .

Therefore $\overline{B C}\neq \overline{D C}$ . Option A is false.

Option B: $\overline{A B}=\overline{D E}$

By CPCTC proved in equation (1) $\overline{AB }=\overline{DE}$ .

Therefore Option B is true.

Option C: $\angle A C B=\angle D E C$

By CPCTC proved in equation (4) $\angle ACB=\angle DCE$ .

Therefore $\angle A C B\neq \angle D E C$ . Option C is false.

Option D: $\angle A B C=\angle E D C$

By CPCTC proved in equation (5) $\angle ABC=\angle DEC$ .

Therefore $\angle A B C\neq \angle E D C$ . Option D is false.

Hence Option B is the correct answer.

$\Rightarrow\overline{A B}=\overline{D E}$

5 0

3 years ago

a rectangle shape whose length is 4 inches less than twice its width. the area is 30 square inches. whats the dimension

Ne4ueva [31]

Using the given information, we can say the following:
Width (W) = x
Length (L) = (2x - 4)
Area of Rectangle = L × W
Using this, we can formulate an equation for rectangle in question:
x(2x - 4) = 30
Expand and move everything to one side:
2x² - 4x - 30 = 0
Simplify by dividing everything on both sides by 2:
x² - 2x - 15 = 0
Factorise:
(x + 5)(x - 3) = 0
Set each factor equal to 0 and solve for x:
x + 5 = 0
x = -5 (a dimension cannot be negative so this is not the solution we want)
x - 3 = 0
x = 3 (this is the solution)
x = W = 3"
L = 2x - 4
L = 2(3) - 4
L = 2"

So, the length the rectangle is 2" and the width is 3"

5 0

3 years ago