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3 years ago

A number x decreased by 3 is 5

Mathematics

1 answer:

zysi [14]3 years ago

5 0

Answer:

Step-by-step explanation:

x - 3 = 5

x = 5 + 3

x = 8

Therefore, the number is 8.

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Check all that apply.<br> x2 + 4x - 9 = 5x + 3

kakasveta [241]

Answer:

The solution set is (-3, 4).

Step-by-step explanation:

x2 + 4x - 9 = 5x + 3

x^2 + 4x - 5x - 9 - 3 = 0

x^2 -x - 12 = 0

(x - 4)(x + 3) = 0

x = -3, 4.

8 0

3 years ago

HELP, please!! What is the area of a trapezoid ABC?

Shtirlitz [24]

It’s c just to let you know I have done this before

3 0

3 years ago

Read 2 more answers

Write the standard form of the equation of the circle with endpoints of a diameter at the points (7,2) and (-9,5).

lys-0071 [83]

Answer:

$(x+1)^2 + (y- 3.5)^2 =64$

Step-by-step explanation:

We are given that endpoints of a diameter at the points (7,2) and (-9,5).

Center of the circle is the mid point of diameter

So, first find the mid point of diameter

Formula : $x=\frac{x_1+x_2}{2} , y=\frac{y_1+y_2}{2}$

$(x_1,y_1)=(7,2)\\(y_1,y_2)=(-9,5)$

Substitute the values in formula

$x=\frac{7-9}{2} , y=\frac{2+5}{2}$

$x=-1 , y=3.5$

So, center of circle = (h,k )=(-1,3.5)

To find length of diameter :

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$(x_1,y_1)=(7,2)\\(y_1,y_2)=(-9,5)$

$d=\sqrt{(-9-7)^2+(5-2)^2}$

$d=\sqrt{265}$

Length of radius = r = $\frac{Diameter}{2}=\frac{\sqrt{256}}{2}$

Standard form of the equation of the circle : $(x -h)^2 + (y- k)^2 = r^2$

(h,k )=(-1,3.5)

$r=\frac{\sqrt{256}}{2}$

Equation of the circle : $(x+1)^2 + (y- 3.5)^2 =(\frac{\sqrt{256}}{2})^2$

Equation of the circle : $(x+1)^2 + (y- 3.5)^2 =64$

Hence the standard form of the equation of the circle with endpoints of a diameter at the points (7,2) and (-9,5) is $(x+1)^2 + (y- 3.5)^2 =64$

5 0

3 years ago

Using the digits 1 to 9, at most one time each, fill in the boxes so that the points make a parallelogram. Mathematically, use p

Lera25 [3.4K]

Answer:

$\left\begin{array}{ccc}A\left( \boxed{3} , \boxed{8} \right)&B\left( \boxed{9} , \boxed{6} \right )\\\\C\left( \boxed{1} , \boxed{4} \right)&D\left( \boxed{7} , \boxed{2} \right )\end{array}\right$

Step-by-step explanation:

The coordinates are:

$\left\begin{array}{ccc}A\left( \boxed{3} , \boxed{8} \right)&B\left( \boxed{9} , \boxed{6} \right )\\\\C\left( \boxed{1} , \boxed{4} \right)&D\left( \boxed{7} , \boxed{2} \right )\end{array}\right$

The parallelogram is attached below.

To verify if these coordinates form a parallelogram, we show that: