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Marat540 [252]
3 years ago
11

Graph the circle X squared plus Y squared -10 X -10 Y +25 equals zero

Mathematics
1 answer:
Natalka [10]3 years ago
8 0

Answer:

25:-8&68___;+you &_356543$"&78

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Carlos packed 3 jars with olives. each jar weighed 14 ounces. what is the total weight of the 3 jars?
nlexa [21]
Then, the total weight would be:   3 * 14 = 42 Ounces
6 0
3 years ago
If 150g of sugar is used for 5 cakes<br> How much is used for 7 cakes?:)
KIM [24]
Well, if 150 grams is used for 5 cakes, we can divide that.
150/5=30. 30 grams of sugar are used for each cake
You want to know how many grams for 7 cakes
30x7=<u><em>210 grams of sugar are used for seven cakes.
</em></u>Hope this helps!<u><em>
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3 0
3 years ago
Read 2 more answers
10. if ABC is equilaterteral, solve for x​
djyliett [7]

Answer:

13

Step-by-step explanation:

If ABC is equilaterteral then all the angles must be equal and 60°

8x - 44 = 60 add 44 to both sides

8x = 104 divide both sides by 8

x = 13

8 0
3 years ago
Please help! ~Find the volume of the triangular prism~
12345 [234]

Answer:

23.04 m^2

Step-by-step explanation:

The volume of a prism is base * height. In the case of a triangular prism, you need to find the area of the base first (aka the triangle).

The area of a triangle is base * height / 2. Since we have the values for those we can substitute the values into the formula and solve.

2.4 * 3.6 / 2 = 3.84.

Now multiply that by height of the prism, which is 6, and we get 23.04 m^2.

Hope this helps!

7 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
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