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earnstyle [38]
3 years ago
11

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
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