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Pavlova-9 [17]
3 years ago
15

The lengths of two sides of a triangle are 12 ft and 13 ft. find the range of possible lengths for the third side

Mathematics
1 answer:
mihalych1998 [28]3 years ago
3 0
Let x be the length of third side.
12 + x > 13
x > 1-------------(1)
13 + x > 12
x > -1 ------------(2)
by combining (1) and (2)
we have x > -1

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A garden table and a bench cost 607 combined. The garden table costs 43 less than the bench. What is the cost of the bench?
Paha777 [63]

Answer is 564

607-40=567

567-3=564


5 0
3 years ago
Read 2 more answers
Please help me withe this proof as well.​
baherus [9]

- D is the midpoint of AB, E is the midpoint of BC

Answer: A. Given

I left off DB||FC because that's not given.  But we can construct it.

Construct line through C parallel to AB.  Extend DE to intersect and call the meet F.

- DB || FC

By Construction

----

- Angle B congruent to angle FCE

Answer: D. Alternate Interior Angles

We have transversal BC across parallel lines AB and CF, so we get congruent angles ABC and FCB aka FCE

- angle BED congruent to angle CEF

Answer: H. Vertical angles are congruent

When we get lines meeting like this we get the usual congruent and supplementary angles.

- Triangle BED congruent to Triangle CEF

Answer: F. Angle Side Angle

We have BE=CE, DBE=FCE, BED=CEF

- DE congruent to FE and DB congruent to FC

Answer: C. CPTCTF

Corresponding parts ...

- AD congruent to DB and DB congruent to FC therefore AD congruent to FC

Answer: E. Transitive Property of Congruent

Things congruent to the same thing are congruent

- ADFC is a parallelogram

Answer: G.  AD and FC are congruent and parallel

Presumably this is a theorem we have already established.

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4 0
3 years ago
Please help me with this !
lubasha [3.4K]

Answer:

One fourth is the same as dividing one whole into four equal fourths. Similarly, dividing 12 by 4 is the same as dividing one whole 12 into fourths.

6 0
2 years ago
For number 6, evaluate the definite integral.
maks197457 [2]
\bf \displaystyle \int\limits_{0}^{28}\ \cfrac{1}{\sqrt[3]{(8+2x)^2}}\cdot dx\impliedby \textit{now, let's do some substitution}\\\\
-------------------------------\\\\
u=8+2x\implies \cfrac{du}{dx}=2\implies \cfrac{du}{2}=dx\\\\
-------------------------------\\\\

\bf \displaystyle \int\limits_{0}^{28}\ \cfrac{1}{\sqrt[3]{u^2}}\cdot \cfrac{du}{2}\implies \cfrac{1}{2}\int\limits_{0}^{28}\ u^{-\frac{2}{3}}\cdot du\impliedby 
\begin{array}{llll}
\textit{now let's change the bounds}\\
\textit{by using } u(x)
\end{array}\\\\
-------------------------------\\\\
u(0)=8+2(0)\implies u(0)=8
\\\\\\
u(28)=8+2(28)\implies u(28)=64

\bf \\\\
-------------------------------\\\\
\displaystyle  \cfrac{1}{2}\int\limits_{8}^{64}\ u^{-\frac{2}{3}}\cdot du\implies \cfrac{1}{2}\cdot \cfrac{u^{\frac{1}{3}}}{\frac{1}{3}}\implies \left. \cfrac{3\sqrt[3]{u}}{2} \right]_8^{64}
\\\\\\
\left[ \cfrac{3\sqrt[3]{(2^2)^3}}{2} \right]-\left[ \cfrac{3\sqrt[3]{2^3}}{2}  \right]\implies \cfrac{12}{2}-\cfrac{6}{2}\implies 6-3\implies 3
3 0
3 years ago
The center of a circle is (4, 6), and an endpoint of a diameter is (2, 5). What is the other endpoint of the diameter?
valentinak56 [21]
Consider this option:
1. if the point (4;6) is the centre of the circle and the point (2;5) is the first endpoint of its diameter, then point (4;6) is the middle point of the diameter (it means that is the middle between the 1st and the 2d endpoints of diameter).
2. using the property described above:
for x of the 2d endpoint of the diameter: x=4*2-2=6;
for y of the 2d endpoint of the diameter: y=6*2-5=7.

answer: (6;7)
3 0
3 years ago
Read 2 more answers
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