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Svetach [21]
3 years ago
8

Find the area of the shaded region please.

Mathematics
1 answer:
Ira Lisetskai [31]3 years ago
7 0
The first shaded triangle is 144 cm. 
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The real numbers $x$ and $y$ are such that \begin{align*} x + y &= 4, \\ x^2 + y^2 &= 22, \\ x^4 &= y^4 - 176 \sqrt{
tamaranim1 [39]

You get everything you need from factoring the last expression:

x^4-y^4=-176\sqrt7

The left side is a difference of squares, and we get another difference of squares upon factoring. We end up with

x^4-y^4=(x^2-y^2)(x^2+y^2)=(x-y)(x+y)(x^2+y^2)

Plug in everything you know and solve for x-y:

-176\sqrt7=(x-y)\cdot4\cdot22\implies x-y=\boxed{-2\sqrt7}

4 0
3 years ago
Read 2 more answers
Fill in the missing work and justification for step 5 when solving 4(x + 1) = 8.
vlada-n [284]

4(x + 1) = 8

4x+4=8

4x=8-4

4x=4

x=4/4

x=1

5 0
3 years ago
Find the angle between the given vectors to the nearest tenth of a degree. u = , v = (2 points)
Liono4ka [1.6K]

Answer:

<h2>3.6°</h2>

Step-by-step explanation:

The question is incomplete. Here is the complete question.

Find the angle between the given vectors to the nearest tenth of a degree.

u = <8, 7>, v = <9, 7>

we will be using the formula below to calculate the angle between the two vectors;

u*v = |u||v| cos \theta

\theta is the angle between the two vectors.

u = 8i + 7j and v = 9i+7j

u*v = (8i + 7j )*(9i + 7j )

u*v = 8(9) + 7(7)

u*v = 72+49

u*v = 121

|u| = √8²+7²

|u| = √64+49

|u| = √113

|v| = √9²+7²

|v| = √81+49

|v| = √130

Substituting the values into the formula;

121= √113*√130 cos θ

cos θ = 121/121.20

cos θ = 0.998

θ = cos⁻¹0.998

θ = 3.6° (to nearest tenth)

Hence, the angle between the given vectors is 3.6°

5 0
3 years ago
HELP PLEASEEE!!!People develop their personal values through_<br> experiences.
Igoryamba
Answer:

Personal experiences.

Explanation:

This is because an individual learn from his/hers interactions and actions within a scenario where he/she can achieve growth.
8 0
3 years ago
Identify the reflection of the figure with vertices E(8,4), F(-16,-8), and G(24,-16) across the y-axis.
Dafna1 [17]

Answer:

\boxed{\text{E$'$(-8, 4), F$'$(16,-8), G$'$(-24,-16)}}

Step-by-step explanation:

When you reflect a point (x, y) in the y-axis, the y-coordinate remains the same, but the x-coordinate gets the opposite sign. Thus,

E (8,4) ⟶ E' (-8,4)

F (-16,-8) ⟶ F' (16,-8)

G (24,-16) ⟶ G' (-24,-16)

\text{The reflected figure has coordinates } \boxed{\textbf{E$'$(-8, 4), F$'$(16,-8), G$'$(-24,-16)}}

The figure EFG and its reflection E'F'G' are shown in the diagram below.

6 0
3 years ago
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