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kondaur [170]
2 years ago
15

What value of d makes the equation true?  d – 34.8 = 5.3

Mathematics
1 answer:
gregori [183]2 years ago
6 0
d -34.8 =5.3\\d - 34.8 + 34.8 = 5.3 + 34.8 \\ d = 5.3 + 34.8\\\\\boxed {d=40.1}
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How to use the GCF and the distributive property of 40+ 50
iren2701 [21]
Hsndmf Nd didjsuduysysbsbsndndhyz
6 0
3 years ago
In 2000, U.S. trade with Saudi Arabia totaled $20.6 billion. Of this total,
xxMikexx [17]

Answer:

30.1%

Step-by-step explanation:

Find the percent that was US exports by dividing 6.2 by 20.6, then multiplying it by 100

(6.2/20.6) x 100

= approximately 30.1

So, approximately 30.1% of the trade was US exports

8 0
2 years ago
Why is the vertex of vertex form y=a(x-h)^2+k (h,k) rather than (-h,k)? If that was y=2(x-5)^2+6, the vertex would be at (5,6) w
Sindrei [870]

9514 1404 393

Answer:

  (5, 6) is (h, k)

Step-by-step explanation:

Vertex form is an instance of the transformation of parent function f(x) = x². It is vertically scaled by a factor of 'a', and translated so the vertex is point (h, k). That is, the transformed vertex is h units right and k units up from that of the parent function (0, 0).

Parent:

  f(x) = x^2

Transformed:

  f(x) = a(x -h)^2 +k

__

When you compare the form to your specific instance, you need to pay attention to what it is that you're comparing. As the attachment shows, ...

  • a = 2
  • -h = -5   ⇒   h = 5
  • k = 6

Hence the vertex is (h, k) = (5, 6). The second attachment shows this on a graph.

7 0
3 years ago
State one form of the Law of Cosines and provide a trick for writing the other two forms and explain when Law of Cosines should
Yuki888 [10]

Solving for <em>Angles</em>

\displaystyle \frac{a^2 + b^2 - c^2}{2ab} = cos∠C \\ \frac{a^2 - b^2 + c^2}{2ac} = cos∠B \\ \frac{-a^2 + b^2 + c^2}{2bc} = cos∠A

* Do not forget to use the <em>inverse</em> function towards the end, or elce you will throw your answer off!

Solving for <em>Edges</em>

\displaystyle b^2 + a^2 - 2ba\:cos∠C = c^2 \\ c^2 + a^2 - 2ca\:cos∠B = b^2 \\ c^2 + b^2 - 2cb\:cos∠A = a^2

You would use this law under <em>two</em> conditions:

  • One angle and two edges defined, while trying to solve for the <em>third edge</em>
  • ALL three edges defined

* Just make sure to use the <em>inverse</em> function towards the end, or elce you will throw your answer off!

_____________________________________________

Now, JUST IN CASE, you would use the Law of Sines under <em>three</em> conditions:

  • Two angles and one edge defined, while trying to solve for the <em>second edge</em>
  • One angle and two edges defined, while trying to solve for the <em>second angle</em>
  • ALL three angles defined [<em>of which does not occur very often, but it all refers back to the first bullet</em>]

* I HIGHLY suggest you keep note of all of this significant information. You will need it going into the future.

I am delighted to assist you at any time.

7 0
2 years ago
Help asap will mark brainliest
valentina_108 [34]

Answer:

the last one

Step-by-step explanation:

its the last one because 1x3 is 3 so you need one with a 3, then 4x2 is 8 and theres half a side so its the last answer with half 3/4 on one side

3 0
3 years ago
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