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3 years ago

How many solutions are there to the following system of equations?

Mathematics

2 answers:

Bas_tet [7]3 years ago

7 0

$-4x-8y=0\\
\underline{-3x+8y=-7}\\
-7x=-7\\
x=1 \Rightarrow B$

sweet [91]3 years ago

6 0

$4x-8y=0 \\
\underline{-3x+8y=-7} \\
4x-3x=0-7 \\
x=-7 \\ \\
4x-8y=0 \\
4 \times (-7)-8y=0 \\
-28-8y=0 \\
-8y=28 \\
y=\frac{28}{-8} \\
y=-\frac{7}{2} \\ \\
 \left \{ {{x=-7} \atop {y=-\frac{7}{2}}} \right.$

There is one solution to this system of equations. The answer is B.

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What is the value of x?

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Answer:

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3 years ago

the length of a boat is 10.8 m Boris buys a scale model of the boat the scale of the model is 1 to 18 work out the length of the

Wittaler [7]

Answer:

The size of the scale model is 60 centimeters.

Step-by-step explanation:

Given that the length of a boat is 10.8 m, and Boris buys a scale model of the boat whose ratio is 1 to 18, to determine the length of the scale model of the boat in centimeters the following calculation must be performed:

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Therefore, the size of the scale model is 60 centimeters.

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2 years ago

A box is in the shape of an equilateral triangular prism. if the box is to be covered with paper on its lateral sides, how many

Zarrin [17]

The total surface area of the triangular prism that has a height of h and the side length of a is given below.

$\rm a(\dfrac{\sqrt3}{2} \ a + 3h)$

<h3>What is a triangular prism?</h3>

A triangular prism is a closed solid that has two parallel triangular bases connected by a rectangle surface.

A box is in the shape of an equilateral triangular prism.

If the box is to be covered with paper on its lateral sides.

Let a be the side length of the equilateral triangle and h be the height of the prism.

Then the surface area of the triangular prism will be

Surface area = 2 × area of triangle + 3 × area of the rectangle

The area of the triangle will be

$\rm Area\ of\ triangle = \dfrac{\sqrt{3}a^2}{4}$

The area of the rectangle will be

$\rm Area \ of \ rectangle = a \ h$

Then the total surface area will be

$\rm Surface\ area = 2 \times \dfrac{\sqrt3 a^2 }{4} + 3 ah\\\\\\Surface\ area = a(\dfrac{\sqrt3}{2} \ a + 3h)$

More about the triangular prism link is given below.

brainly.com/question/21308574

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2 years ago

Sasha is bisecting a segment. First, she places the compass on one endpoint, opens it to a width larger than half of the segment

ycow [4]

Her next step is to repeat the last process of drawing those two arcs. However, they will be mirrored since she swapped endpoints.

Check out the diagram below. Figure 1 is what she already has. Figure 2 is what happens after completing the next step. The red and blue arcs intersect to help form the endpoints of the perpendicular bisector. I used GeoGebra to make the diagrams.

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2 years ago

Find the 2th term of the expansion of (a-b)^4.

vladimir1956 [14]

The second term of the expansion is $-4a^3b$ .

Solution:

Given expression:

$(a-b)^4$

To find the second term of the expansion.

$(a-b)^4$

Using Binomial theorem,

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here, a = a and b = –b

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

Substitute i = 0, we get

$$\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}=1 \cdot \frac{4 !}{0 !(4-0) !} a^{4}=a^4$

Substitute i = 1, we get

$$\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}=\frac{4 !}{3!} a^{3}(-b)=-4 a^{3} b$

Substitute i = 2, we get

$$\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}=\frac{12}{2 !} a^{2}(-b)^{2}=6 a^{2} b^{2}$

Substitute i = 3, we get

$$\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}=\frac{4}{1 !} a(-b)^{3}=-4 a b^{3}$

Substitute i = 4, we get

$$\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}=1 \cdot \frac{(-b)^{4}}{(4-4) !}=b^{4}$

Therefore,

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

$=\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}+\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}+\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}+\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}+\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}$ $=a^{4}-4 a^{3} b+6 a^{2} b^{2}-4 a b^{3}+b^{4}$

Hence the second term of the expansion is $-4a^3b$ .

3 0

3 years ago