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4 years ago

Okay... so I have a math test rn and i need someone who’s really good at math to help!!!

Mathematics

1 answer:

k0ka [10]4 years ago

3 0

Answer:

i am good up to geometry

Step-by-step explanation:

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ABCD is a parallelograms, only find the value of Y?

Aleksandr-060686 [28]

Answer: 9

Step-by-step explanation:

4 0

3 years ago

A gift shop uses two sizes of boxes for presents. These boxes have exactly the same shape. The smaller box is 24 cm long, and th

Vesnalui [34]

Answer:

1,092 cm²

Step-by-step explanation:

Smaller box:

Length = 24 cm

Wrapping paper = 936 cm²

Bigger box:

Length= 28 cm

Wrapping paper = x cm²

Box : wrapping paper = 24 cm : 936 cm²

Box : wrapping paper = 28 cm : x cm²

Equate both ratios to find x

24 : 936 = 28 : x

24/936 = 28/x

Cross product

24 * x = 936 * 28

24x = 26,208

x = 26,208 / 24

x = 1,092 cm²

Bigger box:

Length= 28 cm

Wrapping paper = 1,092 cm²

7 0

3 years ago

Solving for linear systems with two variables <br><br> Y=4x-3<br> 3x-2y=-4

Nimfa-mama [501]

The answer to the question

7 0

3 years ago

Find the coordinates of the circumcenter for ∆DEF with coordinates D(1,3) E (8,3) and F(1,-5). Show your work.

luda_lava [24]

Answer: The coordinates of the circumcenter is $(\frac{9}{2}, -1)$ .

Explanation:

The coordinates of triangle DEF are D(1,3) E (8,3) and F(1,-5).

Distance formula,

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

$DE=\sqrt{(8-1)^2+(3-3)^2}=7$

$FE=\sqrt{(1-8)^2+(-5-3)^2}=\sqrt{7^2+8^2}$

$DF=\sqrt{(1-1)^2+(-5-3)^2}=8$

Since triangle follows pythagoras theorem,

$(DF)^2+(DE)^2=(FE)^2$

Therefore the given triangle is a right angle triangle.

Or plot these points on a coordinate plane. From the figure we can say that the triangle DEF is a right angle triangle.

The circumcenter of a right angle triangle is the midpoint of the hypotenuse.

The hypotenuse is EF. The midpoint of EF is,

$Midpoint =(\frac{8+1}{2}, \frac{3-5}{2} )=(\frac{9}{2}, -1)$

Therefore, the coordinates of the circumcenter is $(\frac{9}{2}, -1)$ .

5 0

3 years ago

The equation of the circle with center (3, -2) and radius 7 is: user: enter in standard form the equation of the line passing th

AURORKA [14]

X^2+y^2-6x+2y-36=0. Use the equation of the circle (x-h)^2+(y-k)^2=radius^2, center (h,k) is (3,-2)

7 0

4 years ago