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2 years ago

Can u solve this puzzle?

Mathematics

1 answer:

finlep [7]2 years ago

4 0

Answer:

i dont think so sorry

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write quadratic equation whose roots are 4 and -3 and whose leading coefficient is 5 use the letter X to represent the variable

AlladinOne [14]

$5(x-4)(x+3)=\\
5(x^2+3x-4x-12)=\\
5(x^2-x-12)=\\
5x^2-5x-60$

5 0

2 years ago

What is the value of X ?

Zielflug [23.3K]

Answer:

x = 12

Step-by-step explanation:

their the same length so plug em equal to each other

x+5 = 2x-7

add 7 to both side to get x+12 = 2x

subtract x from both side to get x = 12

5 0

3 years ago

Read 2 more answers

A bracelet is exactly 6.97 cm across from one side to the other, through the center. What is the approximate area of the bracele

Vadim26 [7]

Answer:

≈ 38.14 cm²

Step-by-step explanation:

Diameter of circle= 6.97 cm

Area of circle = ?

Area = πr² = πd²/4 =
3.14*(6.97)²/4 ≈ 38.14 cm²

3 0

3 years ago

The diameters of a circle is 7m. Find its area to the nearest tenth?? Help please

Lady bird [3.3K]

Area = $\pi r^{2}$

The diameter is double the length of the radius

7 ÷ 2 = 3.5

3.5 × 3.5 = 12.25

$12.25\pi$ = 38.4845100064751

38.4845100064751 rounded to the nearest tenth is 38.5

8 0

2 years ago

Read 2 more answers

A hoop, a uniform solid cylinder, a spherical shell, and a uniform solid sphere are released from rest at the top of an incline.

Serjik [45]

Answer:

Step-by-step explanation:

Given

Hoop, Uniform Solid Cylinder, Spherical shell and a uniform Solid sphere released from Rest from same height

Suppose they have same mass and radius

time Period is given by

$t=\sqrt{\frac{2h}{a}}$ ,where h=height of release

a=acceleration

$a=\frac{g\sin \theta }{1+\frac{I}{mr^2}}$

Where I=moment of inertia

a for hoop

$a=\frac{g\sin \theta }{1+\frac{mr^2}{mr^2}}$

$a=\frac{g\sin \theta }{2}$

a for Uniform solid cylinder

$a=\frac{g\sin \theta }{1+\frac{mr^2}{2mr^2}}$

$a=\frac{2g\sin \theta }{3}$

a for spherical shell

$a=\frac{g\sin \theta }{1+\frac{2mr^2}{3mr^2}}$

$a=\frac{3g\sin \theta }{5}$

a for Uniform Solid

$a=\frac{g\sin \theta }{1+\frac{2mr^2}{5mr^2}}$

$a=\frac{5g\sin \theta }{7}$

time taken will be inversely proportional to the square root of acceleration

$t_1=k\sqrt{2}=1.414k$

$t_2=k\sqrt{\frac{3}{2}}=1.224k$

$t_3=k\sqrt{\frac{5}{3}}=1.2909k$

$t_4=k\sqrt{\frac{7}{5}}=1.183k$

thus first one to reach is Solid Sphere

second is Uniform solid cylinder

third is Spherical Shell

Fourth is hoop

3 0

3 years ago