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3 years ago

Solve for X:-1 Plz help

Mathematics

2 answers:

irina [24]3 years ago

7 0

Answer:

2nd option

Step-by-step explanation:

-1 < x + 3 < 5

Subtract 3 from both sides;

-4 < x < 2

elena55 [62]3 years ago

4 0

Answer:

Lo siento, solo necesitaba los puntos

Step-by-step explanation:

Lo siento, solo necesitaba los puntos

Lo siento, solo necesitaba los puntosvv

Lo siento, solo necesitaba los puntos

Lo siento, solo necesitaba los puntosvvv

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Which expressions are equivalent to -56z + 28.

NeX [460]

Answer:

B and C

Step-by-step explanation:

(-7)•8z = -56z

(-4)•(-7)=28

and

40•(-1.4)= -56z

40•0.7= 28

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2 years ago

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faltersainse [42]

Answer:

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Step-by-step explanation:

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3 years ago

What is the measure of F?<br> 10

kolbaska11 [484]

Answer:

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Step-by-step explanation:

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4 0

3 years ago

Read 2 more answers

Which equation represents a circle that contains the point (-5, 3) and has a center at (-2, 1)? Distance formula: vaa -02 (x - 1

natali 33 [55]

Given:

The center of the circle = (-2,1).

Circle passes through the point (-5,3).

To find:

The equation of the circle.

Solution:

Radius is the distance between the center of the circle and any point on the circle. So, radius of the circle is the distance between the points (-2,1) and (-5,3).

$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$r=\sqrt{(-5-(-2))^2+(3-1)^2}$

$r=\sqrt{(-5+2)^2+(2)^2}$

$r=\sqrt{(-3)^2+(2)^2}$

On further simplification, we get

$r=\sqrt{9+4}$

$r=\sqrt{13}$

The standard form of a circle is:

$(x-h)^2+(y-k)^2=r^2$

Where, (h,k) is the center of the circle and r is the radius of the circle.

Substitute h=-2, k=1 and $r=\sqrt{13}$ .

$(x-(-2))^2+(y-1)^2=(\sqrt{13})^2$

$(x+2)^2+(y-1)^2=13$

Therefore, the equation of the circle is $(x+2)^2+(y-1)^2=13$ .

8 0

3 years ago

Find the missing term and the missing coefficient 6a −( )5a =( ) a2 − 35a

andrew11 [14]

Solving for the missing term and the missing coefficient (6a − )5a = ( ) a^2 − 35a
Let the missing term be X
Let the missing coefficient be Y
Therefore, (6a – X)5a = Y(a^2) – 35a
6a x 5a – X.5a = Y.a^2 – 35a
30a^2 – X.5a = Y.a^2 – 35a
Equating co-efficients,
30a^2 = Y.a^2; X.5a = 35a
30 = Y; 5X = 35
Y = 30; X = 7
Therefore, (6a-7)5a = 30 a^2 – 35a <span>
</span>

3 0

3 years ago