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icang [17]
3 years ago
9

Find the unknown side length

Mathematics
1 answer:
ohaa [14]3 years ago
7 0
You use the formula a² +b² =c²
so 4² +b² =8²
solve for b

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For f(x)=2/(x+1) and g(x)=2x+3, find the following:
pshichka [43]
Im unsure about g(f(4)) though.

7 0
3 years ago
Make y the subject of the formula <br><br> W= x squared - 2yz
AveGali [126]

Answer:

The formula    y = \frac{x^{2} }{2z} + \frac{W}{2z}

Step-by-step explanation:

<em><u>Explanation</u></em>

Given   W = x² - 2 y z

          2 y z = x² + W

Dividing '2z' on both sides , we get

         \frac{2yz}{2z} = \frac{x^{2} + W}{2z}

⇒       y = \frac{x^{2} }{2z} + \frac{W}{2z}

4 0
3 years ago
In a painting, The Great Pyramid of Giza is 45 centimeters tall.
777dan777 [17]

Answer:

The scale is: 1 cm = 3.26 meter

Step-by-step explanation:

Given

Height of Pyramid in Painting = H_p = 45 cm

Actual Height of Pyramid = H_a = 146.5m

In order to find the scale, we can find the ratio between actual height and height in painting.

The ratio can be found by dividing the actual height by height in painting

Scale = \frac{H_a}{H_p}\\= \frac{146.5}{45}\\=3.25555 m/cm

Rounding off to nearest hundredth

= 3.26

This means that the scale is:

1 cm = 3.26 meter

6 0
3 years ago
Use truth tables to show that the following statements are logically equivalent. ∼ P ⇔ Q = (P ⇒∼ Q)∧(∼ Q ⇒ P)
Alina [70]

Answer:  The given logical equivalence is proved below.

Step-by-step explanation:  We are given to use truth tables to show the following logical equivalence :

∼ P ⇔ Q ≡ (P ⇒∼ Q)∧(∼ Q ⇒ P)

We know that

two compound propositions are said to be logically equivalent if they have same corresponding truth values in the truth table.

The truth table is as follows :

P     Q     ∼ P     ∼Q   ∼ P⇔ Q    P ⇒∼ Q    ∼ Q ⇒ P     (P ⇒∼ Q)∧(∼ Q ⇒ P)

T     T         F        F             F            F                   T                       F

T     F         F        T             T             T                   T                       T

F     T         T        F             T            T                   T                       T

F     F         T        T             F            T                   F                       F

Since the corresponding truth vales for ∼ P ⇔ Q and (P ⇒∼ Q)∧(∼ Q ⇒ P) are same, so the given propositions are logically equivalent.

Thus, ∼ P ⇔ Q ≡ (P ⇒∼ Q)∧(∼ Q ⇒ P).

8 0
3 years ago
All else being equal, a study with which of the following error ranges would be
denis-greek [22]

im pretty sure its option B

dont take my word for it

3 0
3 years ago
Read 2 more answers
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