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svetoff [14.1K]
3 years ago
15

Some number added to seventeen

Mathematics
1 answer:
Oksanka [162]3 years ago
5 0

Answer:

x+17

Step-by-step explanation:

x + 17 is the answer

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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
Solve (x + 4)2 – 3(x + 4) – 3 = 0 using substitution. u = Select the solution(s) of the original equation.
bezimeni [28]

Answer:

\dfrac{-5-\sqrt{21}}{2},\ \dfrac{-5+\sqrt{21}}{2}

Step-by-step explanation:

For the equation (x+4)^2-3(x+4)-3=0 use the substitution u=x+4. Then the equation will take look

u^2-3u-3=0.

Solve this quadratic equation:

D=(-3)^2-4\cdot 1\cdot (-3)=9+12=21,\\ \\u_{1,2}=\dfrac{-(-3)\pm \sqrt{21}}{2}=\dfrac{3\pm\sqrt{21}}{2}.

Thus,

x+4=\dfrac{3-\sqrt{21}}{2}\text{ or }x+4=\dfrac{3+\sqrt{21}}{2},\\ \\x_1=\dfrac{3-\sqrt{21}}{2}-4=\dfrac{-5-\sqrt{21}}{2}\text{ or }x_2=\dfrac{3+\sqrt{21}}{2}-4=\dfrac{-5+\sqrt{21}}{2}.

8 0
3 years ago
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1/4 of her original money is 1.25 + 1.60 = 2.85

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The diagram below shows a sandbox and the frame that surrounds it. What is the area of the frame (the shaded area)? (Giving out
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Answer:

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