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nikklg [1K]
3 years ago
15

Please help me ill mark you brainliest and show your work please

Mathematics
1 answer:
Kamila [148]3 years ago
7 0
I cant see what it equals but i’m guessing it is 203(correct me if i’m wrong)

5/8(4x-16)=203
5/8 x 4(x-4)=203
5/2 x (x-4)=203
5/2x-10=203
5x-20=406
5x=406+20
5x=426
x=426/5
x=85 1/5

which is none of the options so you’re gonna have to tell me what it equals to so i can give you the right answer.
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(12x+9)−(3x+7)<br><br><br><br><br> there's a ( sign before 12x
jolli1 [7]
12x-3x=9x 9+7=16

9x+16=25x
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What is leading coefficient of the function f(x) = 3x5 + 6x4 − x − 3?
Morgarella [4.7K]
F(x) = 3x⁵ + 6x⁴ - x - 3

The coefficient of a function f(x) is the number that multiplies the x with the highest exponent., that is 3. Answer a)
3 0
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8 0
4 years ago
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Harman [31]

Answer:

A

Step-by-step explanation:

5 0
3 years ago
Prove diagonals of a rhombus are perpendicular
Lisa [10]

Answer:

A rhombus is a parallelogram with four congruent sides.

So, all sides of rhombus ABCD are congruent.

i.e, \overline{JM} \cong \overline{JK} \cong \overline{KL}\cong \overline{LM}

Also, we know that the diagonals of a parallelogram bisect each other.  

Since a rhombus is a parallelogram.

By property of rhombus , if point N is the intersection of the diagonals as shown in the figure, then

\overline{MN} \cong \overline{NK}         .....[1]

\overline{JN} \cong \overline{NL}

In ΔJNM and ΔJNK

\overline{MN} \cong \overline{NK}     [side]    [by (1)]

\overline{JM} \cong \overline{JK}     [side]          [Given]

By reflexive property states that a segment is congruent to itself:

\overline{JN} \cong \overline{JN}   [Side]                 [Reflexive Property]

SSS(Side-Side-Side) postulates states that if three sides of one triangle are congruent to three sides of another triangle, then the two triangles are congruent.

then by SSS congruence,

\triangle JNM \cong \triangle JNK

By CPCT [Corresponding Part of congruent triangles are congruent]

Since, JNM and JNK are corresponding angles therefore,

\angle JNM \cong JNK

Linear pair theorem states that two angles that form a linear pair are supplementary.

By linear pair theorem, JNM and JNK are supplementary

this mean:

m\angle JNM +m\angle JNK =180^{\circ}

Since, the angles are congruent i.e, \angle JNM \cong JNK

so;

m\angle JNK +m\angle JNK =180^{\circ}

or

m\angle JNK +m\angle JNK =180^{\circ}

2m\angle JNK=180^{\circ}

Simplify:

m\angle JNK =90^{\circ}

also; m\angle JNM =90^{\circ}

therefore, the diagonals of JKLM are perpendicular to each other i,e \overline{JL} \perp \overline{MK}


4 0
3 years ago
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