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3 years ago

What is the missing number in the table?

Mathematics

1 answer:

navik [9.2K]3 years ago

5 0

Answer:

Step-by-step explanation:

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2. In the diagram below, secants PT and PU have been drawn from exterior point P such that the four arcs

scoundrel [369]

Answer:

(a) x = 30°

(b) mRS = 30°

mST = 120°

mTU = 120°

mUR = 90°

Step-by-step explanation:

(a) In the picture attached, the diagram is shown.

Given that m arc RS = x, from the ratios:

m arc ST = 4x

m arc TU = 4x

m arc UR = 3x

The addition of the four arcs must be equal to 360°, then:

x + 4x + 4x + 3x = 360°

12x = 360°

x = 360°/12 = 30°

(b) m arc RS = x = 30°

m arc ST = 4x = 4*30° = 120°

m arc TU = 4x = 4*30° = 120°

m arc UR = 3x = 3*30° = 90°

8 0

3 years ago

It's in the attachment <br> URGENT

maxonik [38]

Step-by-step explanation:

$\frac{2a}{a + b - c} = \frac{2b}{b + c - a} = \frac{2c}{a + c - b} = k \\ \\ by \: theorem \: on \: equal \: ratios \\ \\ \frac{2a + 2b + 2c}{a + b - c + b + c - a + a + c - b} = k \\ \\ \frac{2(a + b + c)}{a + b + c} = k \\ \\ \therefore \: k = 2$

5 0

3 years ago

4. The perimeter of a church window is 60 inches. If the window is in the shape of a regular pentagon, what is the length of eac

mihalych1998 [28]

I think the answer is C

5 0

3 years ago

With problems can be solved

Akimi4 [234]

Answer:

It says Henry scored 9 points every game and he played 4 games. So the equation is 9 × 4.

5 0

3 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

3 years ago