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3 years ago

Could you help me pass 6 garde?

Mathematics

2 answers:

Mice21 [21]3 years ago

8 0

The first one should be B,c,e and the second should be B

morpeh [17]3 years ago

6 0

First picture the answer is B
second picture the answer is B C E

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On a typical day, brianne uses her computer for 3 hours and her hair dryer for 10 minutes. what is the total cost of using both

saw5 [17]

The total cost of using both appliances for 6 days is about 3000-4000 watt.

According to the statement

Brianne uses her computer for 3 hours

Brianne uses hair dryer for 10 minutes

Now we use the formula

C = wtc/1000

substitute the values in it

then

C = 600*6/1000

C = 3.6

and same for hair dryer

So, The total cost of using both appliances for 6 days is about 3000-4000 watt.

Learn more about POWER here

brainly.com/question/12985890

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8 0

2 years ago

A heap of grain is shaped as a cone ADCF with height 5 m and base radius 2 m, as shown on the diagram. A and C are points on the

hoa [83]

Answer:

The angle between [A_F] and the base of the cone = 68.2°

The area of the base of the cone ≈ 12.57 m²

Step-by-step explanation:

The given parameters are;

The height of the cone = 5 m

The base radius of the cone = 2 m

The angle which the A $\hat O$ C = 120°

Therefore, we have;

The angle between [A_F] and the base of the cone = The angle between [CF] and the base of the cone

The angle between [CF] and the base of the cone = tan⁻¹(5/2) = tan⁻¹(2.5) ≈ 68.2°

∴ The angle between [A_F] and the base of the cone = The angle between [CF] and the base of the cone = 68.2°

The angle between [A_F] and the base of the cone = 68.2°

The area of the base of the cone = π × r² = π × 2² = 4·π ≈ 12.57

The area of the base of the cone ≈ 12.57 m².

8 0

3 years ago

Rationalize the denominator.<br> sqrt(3)/(fourth root of 2)

ololo11 [35]

Start with

$\dfrac{\sqrt{3}}{\sqrt[4]{2}}$

Multiply and divide by the cubic root of two:

$\dfrac{\sqrt{3}}{\sqrt[4]{2}}\cdot\dfrac{\sqrt[4]{2^3}}{\sqrt[4]{2^3}}$

Multiplying the fractions leads to

$\dfrac{\sqrt{3}}{\sqrt[4]{2}}\cdot\dfrac{\sqrt[4]{2^3}}{\sqrt[4]{2^3}}=\dfrac{\sqrt{3}\sqrt[4]{2}}{\sqrt[4]{2\cdot 2^3}}=\dfrac{\sqrt{3}\sqrt[4]{2}}{\sqrt[4]{2^4}}=\dfrac{\sqrt{3}\sqrt[4]{2}}{2}$

3 0

3 years ago

What is the solution to this problem?

Nezavi [6.7K]

Answer:

$\frac{2y^2-5y-7}{6y^2+10y+4}-\frac{7}{9y^2+6y}=\frac{6y^2-21y-14}{6y\left(3y+2\right)}$

Step-by-step explanation:

we are given

$\frac{2y^2-5y-7}{6y^2+10y+4}-\frac{7}{9y^2+6y}$

we have to simplify it

Firstly, we will factor it

$2y^2-5y-7=\left(y+1\right)\left(2y-7\right)$

$6y^2+10y+4=2\left(y+1\right)\left(3y+2\right)$

$9y^2+6y=3y(3y+2)$

$=\frac{\left(y+1\right)\left(2y-7\right)}{2\left(y+1\right)\left(3y+2\right)}-\frac{7}{3y(3y+2)}$

now, we can cancel like terms

$=\frac{\left(2y-7\right)}{2\left(3y+2\right)}-\frac{7}{3y(3y+2)}$

now, we can combine it

$=\frac{3y(2y-7\right)}{6y\left(3y+2\right)}-\frac{7\times 2}{6y(3y+2)}$

$=\frac{3y(2y-7\right)-14}{6y\left(3y+2\right)}$

$=\frac{6y^2-21y-14}{6y\left(3y+2\right)}$

so, we get

$=\frac{6y^2-21y-14}{6y\left(3y+2\right)}$

6 0

4 years ago

Select all the possible names for the following angle

Vinil7 [7]

Answer:

The answers are options d and e

3 0

3 years ago