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2 years ago

Pls help 10 pts X = 1 3 7

Mathematics

2 answers:

Black_prince [1.1K]2 years ago

5 0

Answer:

Step-by-step explanation:

Eduardwww [97]2 years ago

3 0

Answer:

x=3

Step-by-step explanation:

Secant-Secant Formula:(whole secant) x (external part) =

(whole secant) x (external part)

4 *(9) = x *(12)

36 =12x

Divide each side by 12

36/12 = 12x/12

3 =x

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Which ratio is higher 15:20 or 12:16. for 35 points

Margarita [4]

To find which ratio is higher, we can first convert the ratios into fractions (that makes it easier, at least for me) and then simplify the fractions and see which one is greater.

Since ratios are basically division, 15:20 = $\frac{15}{20}$ , and

12:16 = $\frac{12}{16}$

Now we simplify these fractions. 15 and 20 have a GCF of 5, so taking out the common number gives us the simplified fraction of $\frac{3}{4}$ .

12 and 16 have a GCF of 4, so taking out the common number gives us the simplified fraction of $\frac{3}{4}$ . Since $\frac{3}{4}$ = $\frac{3}{4}$ , these ratios are the same.

8 0

3 years ago

Please help asap for a test

son4ous [18]

Answer: y=4x−1

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8 0

2 years ago

Read 2 more answers

Solve 7x − 2(x + 1) = 6x + 14. (5 points)<br><br> −12<br><br> 12<br><br> −16<br><br> 16

anygoal [31]

Answer:

X=-16

Step-by-step explanation:

5 0

3 years ago

What are the real and complex solutions of the polynomial equation? x^4-41x^2=-400. *work needed*

pickupchik [31]

<h3>Answer: x = -5, -4, 4 and 5.</h3><h3>All four zeros are real solutions.</h3>

Step-by-step explanation:

Given the polynomial equation $x^4-41x^2=-400$ .

Adding 400 on both sides to get rid 400 from right side and set 0 on right side, we get

$x^4-41x^2+400=-400+400$ .

$x^4-41x^2+400=0$ .

Factoring by product sum rule.

We need product of 400 and sum upto -41.

We can see that 400 = -25 × -16 = 400 and -25-16 = -41.

Therefore,

$x^4-25x^2-16x^2+400=0$

Making it into two groups, we get

$(x^4-25x^2)+(-16x^2+400)=0$

Factoring out GCF of each group, we get

$x^2(x^2-25)-16(x^2-25)=0$

$(x^2-25)(x^2-16) =0$

Factoring out $(x^2-25)$ and $(x^2-16)$ separately by difference of the squares identity $a^2-b^2=(a-b)(a+b)$ , we get

$(x^2-25) = x^2-5^2= (x-5)(x+5)$ and

$x^2-16 = x^2-4^2 =(x-4)(x+4)$ .

Therefore,

$(x-5)(x+5)(x-4)(x+4) =0$

Applying zero product rule,

x-5=0

x+5=0

x-4=0 and

x+4=0.

Therefore,

<h3>x = -5, -4, 4 and 5.</h3><h3>All four zeros are real solutions.</h3>

6 0

3 years ago

Amy wants to find out if the temperature of water affects plant growth. She will water 6 plants with hot water, 6 plants with lu

jek_recluse [69]

Answer:

Temperature of the water

Step-by-step explanation:

In every study, there must be a dependent variable and an independent variable. The effect of the independent variable on the dependent variable is monitored by carefully manipulating the independent variable and observing the response it produces on the dependent variable.

In this study, the independent variable is the temperature of the water while the dependent variable is plant growth. The various temperatures of the water used to water the plant will after its growth.

8 0

3 years ago