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MissTica
3 years ago
9

Which of the following best describes the equation below? y = |x| + 5

Mathematics
1 answer:
sleet_krkn [62]3 years ago
4 0

Answer:

On a graph, you go up 5.

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3x-5)(x2+2x+1) Please show working
Lerok [7]

Answer:

3x^3 + x^2 -7x - 5

Step-by-step explanation:

SOOOOOOO you use the distributive property and multiply by each term.

8 0
3 years ago
Read 2 more answers
Given: AABC is a right triangle.
IgorLugansk [536]

Answer:

19.365

Step-by-step explanation:

The Pythagoreum Theorum tells us that:

a² + b² = c²

where a and b are the legs and c is the hypotenuse.

BC is a leg; we'll lable it a.

AC is the hypotenuse thus labled c.

Plugging their lengths into the above formula will solve for the missing leg length.

5² + x² = 20²

25 + x² = 400

x² = 375

✓x = ✓3 x (5 x 5) x 5

x = 5✓15

x = 19.365

5 0
3 years ago
What is the value of p?
arsen [322]
P=0.6441+0.29=0.9341 (D)
5 0
3 years ago
Fran started practicing her clarinet at 3:30 pm and finished at 5:45 pm. how long did she practice?
yulyashka [42]

Answer:

2 hours and 15 minutes

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
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
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