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Cerrena [4.2K]
3 years ago
5

HELP PLEASE EXPERTS! the question is in the picture

Mathematics
2 answers:
kipiarov [429]3 years ago
7 0
A = 6

Area of a triangle = b•h•1/2

base = 3
height = 4

3•4•1/2 = 6
ahrayia [7]3 years ago
5 0

Answer:

B) 6

Step-by-step explanation:÷

The area of a triangle = 1/2 b* h or \frac{b*h}{2}

Line AB = 4

Line AC = 3

Line BC = x

b = 3

since it is at a right angle,

h = 4

\frac{3*4}{2} = 6

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Which expressions are equivalent to 1.5?
3241004551 [841]

Answer:

B and C

Step-by-step explanation:

We can just solve each to find which expressions are equal to 1.5.

A) 0.15 × 100 = 15

B) 15.0 ÷ 10 = 1.5

C) 0.15 × 10 = 1.5

D) 0.15 ÷ 10 = 0.015

E) 1.5 ÷ 100 = 0.015

From our results, we see that B and C are the only answers that equal 1.5.

6 0
3 years ago
-0.9+28<br> ----<br> 6<br> help me solve the task <br>im a little lazy ​
Anvisha [2.4K]

Answer:

around 4.52.

Fraction form: 4\frac{52}{100}

Step-by-step explanation:

Brainliest?

7 0
3 years ago
Can someone pls answer this?❤️
nata0808 [166]

Answer:

i want to say the answer is A

Step-by-step explanation:

8 0
3 years ago
Use the figure to find the measure of the numbered angles
ivanzaharov [21]

Answer:

angle 2, 4, 6 are 34 degrees

angle 7, 5, 1, 3 are 146 degrees

180-34= 146

5 0
3 years ago
Exercise 3.5. For each of the following functions determine the inverse image of T = {x ∈ R : 0 ≤ [x^2 − 25}.
masya89 [10]

a. The inverse image of f(x) is f⁻¹(x) = ∛(x/3)

b. The inverse image of g(x) is g^{-1}(x) = e^{x}

c. The inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9

<h3 /><h3>The domain of T</h3>

Since T = {x ∈ R : 0 ≤ [x^2 − 25} ⇒ x² - 25 ≥ 0

⇒ x² ≥ 25

⇒ x ≥ ±5

⇒ -5 ≤ x ≤ 5.

<h3>Inverse image of f(x)</h3>

The inverse image of f(x) is f⁻¹(x) = ∛(x/3)

f : R → R defined by f(x) = 3x³

Let f(x) = y.

So, y = 3x³

Dividing through by 3, we have

y/3 = x³

Taking cube root of both sides, we have

x = ∛(y/3)

Replacing y with x we have

y = ∛(x/3)

Replacing y with f⁻¹(x), we have

So,  the inverse image of f(x) is f⁻¹(x) = ∛(x/3)

<h3>Inverse image of g(x)</h3>

The inverse image of g(x) is g^{-1}(x) = e^{x}

g : R+ → R defined by g(x) = ln(x).

Let g(x) = y

y =  ln(x)

Taking exponents of both sides, we have

e^{y} = e^{lnx} \\e^{y} = x

Replacing x with y, we have

y = e^{x}

Replacing y with g⁻¹(x), we have

So, the inverse image of g(x) is g^{-1}(x) = e^{x}

<h3>Inverse image of h(x)</h3>

The inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9

h : R → R defined by h(x) = x − 9

Let y = h(x)

y = x - 9

Adding 9 to both sides, we have

y + 9 = x

Replacing x with y, we have

x + 9 = y

Replacing y with h⁻¹(x), we have

So, the inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9

Learn more about inverse image of a function here:

brainly.com/question/9028678

5 0
3 years ago
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