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2 years ago

15

Can you guys please help me posted picture of question

1 answer:

Dimas [21]2 years ago

3 0

Number of outcomes of rolling a die = 6
Number of outcomes from the spinner = 8

First we have to draw the 6 leaves to represent the 6 outcomes of rolling the die. Next on each of these leaves 8 possible outcomes of the spinner will be drawn using leaves.

So 6 leaves are being divided into further 8 leaves.

Thus total number of leaves will be = 6 x 8 = 48 leaves

Therefore, the correct answer is option D

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Solve the following inequality.<br> -4(1 - 5a) > -124

NemiM [27]

Answer:

a>−6

Step-by-step explanation:

−4(1−5a)>−124

Step 1: Simplify both sides of the inequality.

20a−4>−124

Step 2: Add 4 to both sides.

20a−4+4>−124+4

20a>−120

Step 3: Divide both sides by 20.

20a/20> −120/20

a>−6

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2 years ago

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A percent is a way of a expressing a part out of 100. Remember learning about ratios and how they compare quantities? A part-to-

neonofarm [45]

Answer:A

Step-by-step explanation:

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2 years ago

What is the value of the discriminante for the quadratic equation 0=2x^+x-3

denpristay [2]

Answer:

x1= 1 , x2 = (-1.5)

Step-by-step explanation:

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2 years ago

Triangle $ABC$ has a right angle at $B$. Legs $\overline{AB}$ and $\overline{CB}$ are extended past point $B$ to points $D$ and

Lisa [10]

Answer:

Given :

ABC is a right triangle in which ∠ABC = 90°,

Also, Legs AB and CB are extended past point B to points D and E,

Such that,

$\angle EAC = \angle ACD = 90^{\circ}$

To prove :

$EB\times BD=AB\times BC$

Proof :

In triangles AEC and EBA,

∠EAC= ∠ABE ( right angles )

∠CEA = ∠AEB ( common angles )

By AA similarity postulate,

$\triangle AEC \sim \triangle EBA$ ,

Similarly,

$\triangle AEC \sim \triangle ABC$

$\implies\triangle EBA\sim \triangle ABC-----(1)$

Now, In triangles ADC and CBD,

∠ACD = ∠CBD ( right angles )

∠ADC= ∠BDC ( common angles )

By AA similarity postulate,

$\triangle ADC \sim \triangle CBD$ ,

Similarly,

$\triangle ADC \sim \triangle ABC$

$\implies \triangle CBD\sim \triangle ABC-----(2)$

From equations (1) and (2),

$\triangle EBA\sim \triangle CBD$

The corresponding sides of similar triangles are in same proportion,

$\frac{EB}{BC}=\frac{AB}{BD}$

$\implies EB\times BD=AB\times BC$

Hence, proved....

5 0

2 years ago